™Let the relation of knowledge to real life be very visible to your pupils and let them understand how by knowledge the world could be transformed. – BERTRAND RUSSELL • 11.1 Introduction In the preceding Chapter 10, we have studied various forms of the equations of a line. In this Chapter, we shall study about some other curves, viz., circles, ellipses, parabolas and hyperbolas. The names parabola and hyperbola are given by Apollonius. These curves are in fact, known as conic sections or more commonly conics because they can be obtained as intersections of a plane with a double napped right circular cone. These curves have a very wide range of applications in fields such as planetary motion, design of telescopes and antennas, reflectors in flashlights and automobile headlights, etc. Now, in the subsequent sections we will see how the intersection of a plane with a double napped right circular cone results in different types of curves. 11.2 Sections of a Cone Let l be a fixed vertical line and m be another line intersecting it at a fixed point V and inclined to it at an angle α (Fig11.1). Suppose we rotate the line m around the line l in such a way that the angle α remains constant. Then the surface generated is a double-napped right circular hollow cone herein after referred as Fig 11. 1 cone and extending indefinitely far in both directions (Fig11.2). The point V is called the vertex; the line l is the axis of the cone. The rotating line m is called a generator of the cone. The vertex separates the cone into two parts called nappes. If we take the intersection of a plane with a cone, the section so obtained is called a conic section. Thus, conic sections are the curves obtained by intersecting a right circular cone by a plane. We obtain different kinds of conic sections depending on the position of the intersecting plane with respect to the cone and by the angle made by it with the vertical axis of the cone. Let β be the angle made by the intersecting plane with the vertical axis of the cone (Fig11.3). The intersection of the plane with the cone can take place either at the vertex of the cone or at any other part of the nappe either below or above the vertex. 11.2.1 Circle, ellipse, parabola and hyperbola When the plane cuts the nappe (other than the vertex) of the cone, we have the following situations: (a) When β = 90o, the section is a circle (Fig11.4). (b) When α < β < 90o, the section is an ellipse (Fig11.5). (c) When β = α; the section is a parabola (Fig11.6). (In each of the above three situations, the plane cuts entirely across one nappe of the cone). (d) When 0 ≤β < α; the plane cuts through both the nappes and the curves of intersection is a hyperbola (Fig11.7). Fig 11. 4 Fig 11. 5 Fig 11. 7 11.2.2 Degenerated conic sections When the plane cuts at the vertex of the cone, we have the following different cases: (a) When α < β≤ 90o, then the section is a point (Fig11.8). (b) When β = α, the plane contains a generator of the cone and the section is a straight line (Fig11.9). It is the degenerated case of a parabola. (c) When 0 ≤β < α, the section is a pair of intersecting straight lines (Fig11.10). It is the degenerated case of a hyperbola. In the following sections, we shall obtain the equations of each of these conic sections in standard form by defining them based on geometric properties. Fig 11. 10 11.3 Circle Definition 1 A circle is the set of all points in a plane that are equidistant from a fixed point in the plane. The fixed point is called the centre of the circle and the distance from the centre to a point on the circle is called the radius of the circle (Fig 11.11). Fig 11. 12 Fig 11. 11 The equation of the circle is simplest if the centre of the circle is at the origin. However, we derive below the equation of the circle with a given centre and radius (Fig 11.12). Given C (h, k) be the centre and rthe radius of circle. Let P(x, y) be any point on the circle (Fig11.12). Then, by the definition, | CP | = r. By the distance formula, we have (x–h)2 + (y–k )2 = r i.e. (x– h)2 + (y– k)2 = r2 This is the required equation of the circle with centre at (h,k) and radius r. Example 1 Find an equation of the circle with centre at (0,0) and radius r. Solution Here h= k= 0. Therefore, the equation of the circle is x2 + y2 = r2. Example 2 Find the equation of the circle with centre (–3, 2) and radius 4. Solution Here h= –3, k= 2 and r= 4. Therefore, the equation of the required circle is (x+ 3)2 + (y–2)2 = 16 Example 3 Find the centre and the radius of the circle x2 + y2 + 8x+ 10y– 8 = 0 Solution The given equation is (x2 + 8x) + (y2 + 10y) = 8 Now, completing the squares within the parenthesis, we get (x2 + 8x+ 16) + (y2 + 10y+ 25) = 8 + 16 + 25 i.e. (x+ 4)2 + (y+ 5)2 = 49 i.e. {x– (– 4)}2 + {y– (–5)}2 = 72 Therefore, the given circle has centre at (– 4, –5) and radius 7. Example 4 Find the equation of the circle which passes through the points (2, – 2), and (3,4) and whose centre lies on the line x + y = 2. Solution Let the equation of the circle be (x – h)2 + (y – k)2 = r2. Since the circle passes through (2, – 2) and (3,4), we have (2 – h)2 + (–2 – k)2 = r2 ... (1) and (3 – h)2 + (4 – k)2 = r2 ... (2) Also since the centre lies on the line x + y = 2, we have h + k = 2 ... (3) Solving the equations (1), (2) and (3), we get h = 0.7, k = 1.3 and r2 = 12.58 Hence, the equation of the required circle is (x – 0.7)2 + (y – 1.3)2 = 12.58. EXERCISE 11.1 In each of the following Exercises 1 to 5, find the equation of the circle with 1. centre (0,2) and radius 2 2. centre (–2,3) and radius 4 11 13. centre ( , ) and radius 4. centre (1,1) and radius 224 12 5. centre (–a, –b) and radius In each of the following Exercises 6 to 9, find the centre and radius of the circles. 6. (x + 5)2 + (y – 3)2 = 36 7. x2 + y2 – 4x – 8y – 45 = 0 8. x2 + y2 – 8x + 10y – 12 = 0 9. 2x2 + 2y2 – x = 0 10. Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x + y = 16. 11. Find the equation of the circle passing through the points (2,3) and (–1,1) and whose centre is on the line x – 3y – 11 = 0. 12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3). 13. Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes. 14. Find the equation of a circle with centre (2,2) and passes through the point (4,5). 15. Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25? 11.4 Parabola Definition 2 A parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point (not on the line) in the plane. The fixed line is called the directrix of the parabola and the fixed point F is called the focus (Fig 11.13). (‘Para’ means ‘for’ and ‘bola’ means ‘throwing’, i.e., the shape described when you throw a ball in the air). A line through the focus and perpendicular to the directrix is called the axis of the parabola. The point of intersection of parabola with the axis is called the vertex of the parabola (Fig11.14). 11.4.1 Standard equations of parabola The equation of a parabola is simplest if the vertex is at the origin and the axis of symmetry is along the x-axis or y-axis. The four possible such orientations of parabola are shown below in Fig11.15 (a) to (d). We will derive the equation for the parabola shown above in Fig 11.15 (a) with focus at (a, 0) a > 0; and directricx x = – a as below: Let F be the focus and l the directrix. Let FM be perpendicular to the directrix and bisect FM at the point O. Produce MO to X. By the definition of parabola, the mid-point O is on the parabola and is called the vertex of the parabola. Take O as origin, OX the x-axis and OY perpendicular to it as the y-axis. Let the distance from the directrix to the focus be 2a. Then, the coordinates of the focus are (a, 0), and the equation of the directrix is x + a = 0 as in Fig11.16. Let P(x, y) be any point on the parabola such that PF = PB, where PB is perpendicular to l. The coordinates of B are (– a, y). By the distance formula, we have PF = 2 2 and PB = 2(x–a) + y(x + a) Since PF = PB, we have 22 2(x–a) + y = (x + a) i.e. (x – a)2 + y2 = (x + a)2 or x2 – 2ax + a2 + y2 = x2 + 2ax + a2 or y2 = 4ax ( a > 0). Hence, any point on the parabola satisfies y2 = 4ax. ... (2) Conversely, let P(x, y) satisfy the equation (2) PF =(x–a)2 + y2= (x–a)2 + 4ax = PB ... (3) and so P(x,y) lies on the parabola. Thus, from (2) and (3) we have proved that the equation to the parabola with vertex at the origin, focus at (a,0) and directrix x = – a is y2 = 4ax. Discussion In equation (2), since a > 0, x can assume any positive value or zero but no negative value and the curve extends indefinitely far into the first and the fourth quadrants. The axis of the parabola is the positive x-axis. Similarly, we can derive the equations of the parabolas in: Fig 11.15 (b) as y2 = – 4ax, Fig 11.15 (c) as x2 = 4ay, Fig 11.15 (d) as x2 = – 4ay, These four equations are known as standard equations of parabolas. axis; vertex at the origin and thereby the directrix is parallel to the other coordinate axis. However, the study of the equations of parabolas with focus at any point and any line as directrix is beyond the scope here. From the standard equations of the parabolas, Fig11.15, we have the following observations: 1. Parabola is symmetric with respect to the axis of the parabola.If the equation has a y2 term, then the axis of symmetry is along the x-axis and if the equation has an x2 term, then the axis of symmetry is along the y-axis. 2. When the axis of symmetry is along the x-axis the parabola opens to the (a) right if the coefficient of x is positive, (b) left if the coefficient of x is negative. 3. When the axis of symmetry is along the y-axis the parabola opens (c) upwards if the coefficient of y is positive. (d) downwards if the coefficient of y is negative. 11.4.2 Latus rectum Definition 3 Latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus and whose end points lie on the parabola (Fig11.17). To find the Length of the latus rectum of the parabola y2 = 4ax (Fig 11.18). By the definition of the parabola, AF = AC. But AC= FM = 2a Hence AF= 2a. And since the parabola is symmetric with respect to x-axis AF = FB and so AB = Length of the latus rectum = 4a. Example 5 Find the coordinates of the focus, axis, the equation of the directrix and latus rectum of the parabola y2 = 8x. Solution The given equation involves y2, so the axis of symmetry is along the x-axis. The coefficient of x is positive so the parabola opens to the right. Comparing with the given equation y2 = 4ax, we find that a = 2. Thus, the focus of the parabola is (2, 0) and the equation of the directrix of the parabola is x = – 2 (Fig 11.19). Length of the latus rectum is 4a = 4 × 2 = 8. Example 6 Find the equation of the parabola with focus (2,0) and directrix x = – 2. Solution Since the focus (2,0) lies on the x-axis, the x-axis itself is the axis of the parabola. Hence the equation of the parabola is of the form either y2 = 4ax or y2 = – 4ax. Since the directrix is x = – 2 and the focus is (2,0), the parabola is to be of the form y2 = 4ax with a = 2. Hence the required equation isy2 = 4(2)x = 8x Example 7 Find the equation of the parabola with vertex at (0, 0) and focus at (0, 2). Solution Since the vertex is at (0,0) and the focus is at (0,2) which lies on y-axis, the y-axis is the axis of the parabola. Therefore, equation of the parabola is of the form x2 = 4ay. thus, we have x2 = 4(2)y, i.e., x2 = 8y. Example 8 Find the equation of the parabola which is symmetric about the y-axis, and passes through the point (2,–3). Solution Since the parabola is symmetric about y-axis and has its vertex at the origin, the equation is of the form x2 = 4ay or x2 = – 4ay, where the sign depends on whether the parabola opens upwards or downwards. But the parabola passes through (2,–3) which lies in the fourth quadrant, it must open downwards. Thus the equation is of the form x2 = – 4ay. Since the parabola passes through ( 2,–3), we have 122 = – 4a (–3), i.e., a = 3 Therefore, the equation of the parabola is 1 x2 = −4⎛⎞⎜⎟y, i.e., 3x2 = – 4y.3⎝⎠EXERCISE 11.2 In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum. 1. y2 = 12x 2. x2 = 6y 3. y2 = – 8x 4. x2 = – 16y 5. y2 = 10x 6. x2 = – 9y In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions: 7. Focus (6,0); directrix x = – 6 8. Focus (0,–3); directrix y = 3 9. Vertex (0,0); focus (3,0) 10. Vertex (0,0); focus (–2,0) 11. Vertex (0,0) passing through (2,3) and axis is along x-axis. 12. Vertex (0,0), passing through (5,2) and symmetric with respect to y-axis. 11. 5 Ellipse Definition 4 An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant. The two fixed points are called the foci (plural of ‘focus’) of the ellipse (Fig11.20). Fig 11.20 The mid point of the line segment joining the foci is called the centre of the ellipse. The line segment through the foci of the ellipse is called the major axis and the line segment through the centre and perpendicular to the major axis is called the minor axis. The end points of the major axis are called the vertices of the ellipse(Fig 11.21). We denote the length of the major axis by 2a, the length of the minor axis by 2b and the distance between the foci by 2c. Thus, the length of the semi major axis is a and semi-minor axis is b (Fig11.22). 11.5.1 Relationship between semi-major axis, semi-minor axis and the distance of the focus from the centre of the ellipse (Fig 11.23). Take a point P at one end of the major axis. Sum of the distances of the point P to the foci is FP + FP = FO + OP + FP1 212(Since, F1P = F1O + OP) = c + a + a – c = 2a Fig 11.23 Take a point Q at one end of the minor axis. Sum of the distances from the point Q to the foci is 2222 22F1Q + F2Q= b +c + b + c = 2 b + c Since both P and Q lies on the ellipse. By the definition of ellipse, we have 22 222 b + c = 2a, i.e., a = b + c or a2= b2 + c2 , i.e., c = 11.5.2 Special cases of an ellipse In the equation c2 = a2 – b2 obtained above, if we keep a fixed and vary c from 0 to a, the resulting ellipses will vary in shape. Case (i) When c = 0, both foci merge together with the centre of the ellipse and a2 = b2, i.e., a = b, and so the ellipse becomes circle (Fig11.24). Thus, circle is a special case of an ellipse which is dealt in Section 11.3. Fig 11.24 Case (ii) When c = a, then b = 0. The ellipse reduces to the line segment F1F2 joining the two foci (Fig11.25). 11.5.3 Eccentricity Fig 11.25 Definition 5 The eccentricity of an ellipse is the ratio of the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse (eccentricity is denoted by e) i.e., e = c . a Then since the focus is at a distance of c from the centre, in terms of the eccentricity the focus is at a distance of ae from the centre. 11.5.4 Standard equations of an ellipse The equation of an ellipse is simplest if the centre of the ellipse is at the origin and the foci are (a) Fig 11.26on the x-axis or y-axis. The two such possible orientations are shown in Fig 11.26. We will derive the equation for the ellipse shown above in Fig 11.26 (a) with foci on the x-axis. Let F1 and F2 be the foci and O be the midpoint of the line segment F1F2. Let O be the origin and the line from O through F2 be the positive x-axis and that through F1as the negative x-axis. Let, the line through O perpendicular to the x-axis be the y-axis. Let the coordinates of F1 be (– c, 0) and F2 be (c, 0) (Fig 11.27). Let P(x, y) be any point on the ellipse such that the sum of the distances from P to the two x2 y2 +=1foci be 2a so given 2 2ab PF1 + PF2 = 2a. ... (1) Using the distance formula, we have Fig 11.27 22 22(x +c) + y + (x − c) + y = 2a )22 )22i.e., (x +c + y = 2a – (x − c + y Squaring both sides, we get 22 22(x + c)2 + y2 = 4a2 – 4a (x −c) + y + (x −c) + ywhich on simplification gives c)2 2(x − c + y = a − x a Squaring again and simplifying, we get x2 y2 2 +2 2 = 1 aa − c x2 y2 i.e., 2 +2 = 1 (Since c2 = a2 – b2)ab Hence any point on the ellipse satisfies x2 y2 + = 1. ... (2) a2 b2 Conversely, let P (x, y) satisfy the equation (2) with 0 < c < a. Then ⎛ x2 ⎞ y2 = b2 ⎜⎜ 1 − 2 ⎟⎟ a⎝⎠ Therefore, PF1 = (x + c)2 + y2 )2 2 ⎛ a2 − x2 ⎞ = (x + c + b ⎜⎜ 2 ⎟⎟ a⎝⎠ 2 22 ⎛ a2 − x2 ⎞ = (xc)(a −c )⎜ 2 ⎟ (since b2 = a2 – c2)++ a⎝⎠⎛ cx ⎞2 c a +=a + x= ⎜⎟⎝ a ⎠ a c Similarly PF2a − x= a cc ... (3)Hence PF1 + PF2 = a + x + a– x = 2a aa 22xySo, any point that satisfies 2 + 2 = 1, satisfies the geometric condition and soab P(x, y) lies on the ellipse. Hence from (2) and (3), we proved that the equation of an ellipse with centre of the origin and major axis along the x-axis is x2 y2 + = 1. a2 b2 Discussion From the equation of the ellipse obtained above, it follows that for every point P (x, y) on the ellipse, we have 22x = 1 − y ≤ 1, i.e., x2 ≤ a2, so – a ≤ x ≤ a. a2 b2 Therefore, the ellipse lies between the lines x = – a and x = a and touches these lines. Similarly, the ellipse lies between the lines y = – b and y = b and touches these lines. x2 y2 Similarly, we can derive the equation of the ellipse in Fig 11.26 (b) as 2 + 2 =1.ba These two equations are known as standard equations of the ellipses. From the standard equations of the ellipses (Fig11.26), we have the following observations: 1. Ellipse is symmetric with respect to both the coordinate axes since if (x, y) is a point on the ellipse, then (– x, y), (x, –y) and (– x, –y) are also points on the ellipse. 2. The foci always lie on the major axis. The major axis can be determined by finding the intercepts on the axes of symmetry. That is, major axis is along the x-axis if the coefficient of x2 has the larger denominator and it is along the y-axis if the coefficient of y2 has the larger denominator. 11.5.5 Latus rectum Definition 6 Latus rectum of an ellipse is a line segment perpendicular to the major axis through any of the foci and whose end points lie on the ellipse (Fig 11.28). To find the length of the latus rectum x2 y2 of the ellipse 2+ 2=1 ab Let the length of AF2 be l. Fig 11. 28 Then the coordinates of A are (c, l ),i.e., (ae, l ) x2 y2 Since A lies on the ellipse a2 + b2 =1, we have ()2 l2ae 2 + 2 =1 ab ⇒ l2 = b2 (1 – e2) 222 2c a–b bBut e2 == =1 – 22 2aa a b4 b2 Therefore l2 = 2 , i.e., l = aa Since the ellipse is symmetric with respect to y-axis (of course, it is symmetric w.r.t. 2b2 both the coordinate axes), AF2 = F2B and so length of the latus rectum is . a Example 9 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse x2 y2 +=1 25 9 x2 y2 Solution Since denominator of is larger than the denominator of , the major25 9 x2 y2 axis is along the x-axis. Comparing the given equation with a2 + b2 =1, we get a= 5 and b = 3. Also Therefore, the coordinates of the foci are (– 4,0) and (4,0), vertices are (– 5, 0) and (5, 0). Length of the major axis is 10 units length of the minor axis 2bis 6 units and the 4 2b2 18 eccentricity is and latus rectum is = .5 a 5 Example 10 Find the coordinates of the foci, the vertices, the lengths of major and minor axes and the eccentricity of the ellipse 9x2 + 4y2 = 36. Solution The given equation of the ellipse can be written in standard form as x2 y2 +=1 49 y2 x2 Since the denominator of is larger than the denominator of , the major axis is9 4 along the y-axis. Comparing the given equation with the standard equation x2 y2 b2 + a2 =1, we have b= 2 and a= 3. a 3 Hence the foci are (0, 5 ) and (0, – 5 ), vertices are (0,3) and (0, –3), length of the major axis is 6 units, the length of the minor axis is 4 units and the eccentricity of the 5ellipse is .3 Example 11 Find the equation of the ellipse whose vertices are (± 13, 0) and foci are (± 5, 0). Solution Since the vertices are on x-axis, the equation will be of the form x 2 y2 2 +2 =1, where ais the semi-major axis.ab Given that a = 13, c = ± 5. Therefore, from the relation c2 = a2 – b2, we get 25 =169 – b2 , i.e., b = 12 x2 y2 Hence the equation of the ellipse is +=1.169 144 Example 12 Find the equation of the ellipse, whose length of the major axis is 20 and foci are (0, ± 5). Solution Since the foci are on y-axis, the major axis is along the y-axis. So, equation x2 y2 of the ellipse is of the form 2 + 2 =1.ba Given that a = semi-major axis = 20 =10 2 and the relation c2= a2 – b2 gives 52 = 102 – b2 i.e.,b2 = 75 Therefore, the equation of the ellipse is x2 y2 +=1 75 100 Example 13 Find the equation of the ellipse, with major axis along the x-axis and passing through the points (4, 3) and (– 1,4). x2 y2 Solution The standard form of the ellipse is + = 1. Since the points (4, 3) a2 b2 and (–1, 4) lie on the ellipse, we have 16 9+= 1 ... (1) a2 b2 1 16 and 2 +2 = 1 ….(2)ab 2 247 2 247Solving equations (1) and (2), we find that a = and b = .7 15 Hence the required equation is x2 y2 +=1 ⎛ 247 ⎞ 247 , i.e., 7x2 + 15y2 = 247. ⎜⎟⎝ 7 ⎠ 15 EXERCISE 11.3 In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse. 2 2 2222x y xyxy1. +=1 2. +=1 3. +=1 36 16 425 169 2 2 2222x y xyxy4. +=1 5. +=1 6. + = 125 100 49 36 100 400 7. 36x2 + 4y2 = 144 8. 16x2 + y2 = 16 9. 4x2 + 9y2 = 36 In each of the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions: 10. Vertices (± 5, 0), foci (± 4, 0) 11. Vertices (0, ± 13), foci (0, ± 5) 12. Vertices (± 6, 0), foci (± 4, 0) 13. Ends of major axis (± 3, 0), ends of minor axis (0, ± 2) 14. Ends of major axis (0, ± 5 ), ends of minor axis (± 1, 0) 15. Length of major axis 26, foci (± 5, 0) 16. Length of minor axis 16, foci (0, ± 6). 17. Foci (± 3, 0), a = 4 18. b = 3, c = 4, centre at the origin; foci on the x axis. 19. Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1,6). 20. Major axis on the x-axis and passes through the points (4,3) and (6,2). 11.6 Hyperbola Definition 7 A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant. Fig 11.29 The term “difference” that is used in the definition means the distance to the farther point minus the distance to the closer point. The two fixed points are called the foci of the hyperbola. The mid-point of the line segment joining the foci is called the centre of the hyperbola. The line through the foci is called the transverse axis and the line through the centre and perpendicular to the transverse axis is called the conjugate axis. The points at which the hyperbola intersects the transverse axis are called the vertices of the hyperbola (Fig 11.29). We denote the distance between the two foci by 2c, the distance between two vertices (the length of the transverse axis) by 2a and we define the quantity b as 22b = c–a Also 2b is the length of the conjugate axis Fig 11.30 (Fig 11.30). To find the constant P1F2 – P1:F1 By taking the point P at A and B in the Fig 11.30, we have – BF= AF – AF (by the definition of the hyperbola)BF1 2 21BA +AF1– BF2 = AB + BF2– AF1 i.e., AF1 = BF2 So that,BF – BF= BA + AF– BF = BA = 2a12 1211.6.1 Eccentricity cDefinition 8 Just like an ellipse, the ratio e = is called the eccentricity of the ahyperbola. Since c ≥ a, the eccentricity is never less than one. In terms of the eccentricity, the foci are at a distance of ae from the centre. 11.6.2 Standard equation of Hyperbola The equation of a hyperbola is simplest if the centre of the hyperbola is at the origin and the foci are on the x-axis or y-axis. The two such possible orientations are shown in Fig11.31. We will derive the equation for the hyperbola shown in Fig 11.31(a) with foci on the x-axis. Let F1 and F2 be the foci and O be the mid-point of the line segment F1F2. Let O be the origin and the line through O through F2 be the positive x-axis and that through F1 as the negative x-axis. The line through O perpendicular to the x-axis be the y-axis. Let the coordinates of F1 be (– c,0) and F2 be (c,0) (Fig 11.32). Let P(x, y) be any point on the hyperbola such that the difference of the distances from P to the farther point minus the closer point be 2a. So given, PF1 – PF2 = 2a Using the distance formula, we have 22 22(x + c) + y– (x–c ) + y = 2a 22 22i.e., (x + c) + y = 2a + (x–c ) + y Squaring both side, we get (x + c)2 + y2 = 4a2 + 4a (x–c)2 + y2 +(x – c)2 + y2 and on simplifying, we get cx– a = (x–c)2 + y2 a On squaring again and further simplifying, we get x2 y2 – =12 22a c–a x2 y2 i.e., a2 –b2 =1 (Since c2 – a2 = b2) x2 y2 Hence any point on the hyperbola satisfies 2 – 2 =1 ab Conversely, let P(x, y) satisfy the above equation with 0 < a < c. Then ⎛ x2 –a2 ⎞ y2 = b2 ⎜ 2 ⎟ a⎝⎠ Therefore, PF1 = + 2 2(x + c) + y ⎛ 22 ⎞x–a c xc= + ( + )2 + b2 ⎜ 2 ⎟ = a + x a⎝⎠a aSimilarly, PF2 = a – x c cIn hyperbola c > a; and since P is to the right of the line x = a, x > a, x > a. Therefore,a cc a – x becomes negative. Thus, PF2 = x – a. aa c cxTherefore PF1 – PF2 = a + x – + a = 2a aaAlso, note that if P is to the left of the line x = – a, then ⎛ c ⎞ c =–a xPF1 ⎜+ ⎟, PF2 = a – x .⎝ a ⎠a x2 y2 In that case P F– PF = 2a. So, any point that satisfies – =1, lies on the2 1 22a b hyperbola. Thus, we proved that the equation of hyperbola with origin (0,0) and transverse axis x2 y2 along x-axis is – =1. a2 b2 $Note A hyperbola in which a = b is called an equilateral hyperbola. Discussion From the equation of the hyperbola we have obtained, it follows that, we x2 y2 have for every point (x, y) on the hyperbola, 1 ≥ 1.=+ a2 b2 x ≥ 1, i.e., x ≤ – a or x ≥ a. Therefore, no portion of the curve lies between thei.e, a lines x = + a and x = – a, (i.e. no real intercept on the conjugate axis). 2 2y xSimilarly, we can derive the equation of the hyperbola in Fig 11.31 (b) as 2 −2 = 1 a b These two equations are known as the standard equations of hyperbolas. $Note The standard equations of hyperbolas have transverse and conjugate axes as the coordinate axes and the centre at the origin. However, there are hyperbolas with any two perpendicular lines as transverse and conjugate axes, but the study of such cases will be dealt in higher classes. From the standard equations of hyperbolas (Fig11.29), we have the following observations: 1. Hyperbola is symmetric with respect to both the axes, since if (x, y) is a point on the hyperbola, then (– x, y), (x, – y) and (– x, – y) are also points on the hyperbola. 2. The foci are always on the transverse axis. It is the positive term whose x2 y2 denominator gives the transverse axis. For example, – =1 9 16 y2 x2 has transverse axis along x-axis of length 6, while – =1 25 16 has transverse axis along y-axis of length 10. 11.6.3 Latus rectum Definition 9 Latus rectum of hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose end points lie on the hyperbola. 2b2 As in ellipse, it is easy to show that the length of the latus rectum in hyperbola is . a Example 14 Find the coordinates of the foci and the vertices, the eccentricity,the length of the latus rectum of the hyperbolas: x2 y2 (i) – =1 , (ii) y2 – 16x2 = 169 16 x2 y2 Solution (i) Comparing the equation – =1 with the standard equation9 16 x2 y2 2 – 2 =1 ab Here, a = 3, b = 4 and c = Therefore, the coordinates of the foci are (± 5, 0) and that of vertices are (± 3, 0).Also, c 52b2 32 The eccentricity e = = . The latus rectum == a 3 a 3 y2 x2 (ii) Dividing the equation by 16 on both sides, we have – =1 16 1 y2 x2 Comparing the equation with the standard equation a2 –b2 =1, we find that a = 4, b = 1 and c = a2 + b2 = 16 +1 = 17 . Therefore, the coordinates of the foci are (0, ± 17 ) and that of the vertices are (0, ± 4). Also, 22c 17 b 1The eccentricity e == . The latus rectum == . a 4 a 2 Example 15 Find the equation of the hyperbola with foci (0, ± 3) and vertices 11(0, ± ).2 Solution Since the foci is on y-axis, the equation of the hyperbola is of the form y 2 x2 2 – 2 =1 ab 11 11Since vertices are (0, ± ), a = 22 25Also, since foci are (0, ± 3); c = 3 and b2 = c2 – a2 = .4 Therefore, the equation of the hyperbola is 22yx – = 1, i.e., 100 y2 – 44 x2 = 275.11 25⎛⎞⎛ ⎞ ⎜⎟⎜ ⎟44⎝⎠⎝ ⎠Example 16 Find the equation of the hyperbola where foci are (0, ±12) and the length of the latus rectum is 36. Solution Since foci are (0, ± 12), it follows that c = 12. 2b2 Length of the latus rectum = =36 or b2 = 18a a Therefore c2 = a2 + b2; gives 144 = a2 + 18a i.e., a2 + 18a – 144 = 0, So a = – 24, 6. Since a cannot be negative, we take a = 6 and so b2 = 108. y2 x2 Therefore, the equation of the required hyperbola is – =1, i.e., 3y2 – x2 = 10836 108 EXERCISE 11.4 In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas. 22 22xy yx1. – =1 2. – =1 3. 9y2 – 4x2 = 36169 927 4. 16x2 – 9y2 = 576 5. 5y2 – 9x2 = 36 6. 49y2 – 16x2 = 784. In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions. 7. Vertices (± 2, 0), foci (± 3, 0) 8. Vertices (0, ± 5), foci (0, ± 8) 9. Vertices (0, ± 3), foci (0, ± 5) 10. Foci (± 5, 0), the transverse axis is of length 8. 11. Foci (0, ±13), the conjugate axis is of length 24. 12. Foci (± 3 5 , 0), the latus rectum is of length 8. 13. Foci (± 4, 0), the latus rectum is of length 12 14. vertices (± 7,0), e = .43 15. Foci (0, ± 10 ), passing through (2,3) Miscellaneous Examples Example 17 The focus of a parabolic mirror as shown in Fig 11.33 is at a distance of 5 cm from its vertex. If the mirror is 45 cm deep, find the distance AB (Fig 11.33). Solution Since the distance from the focus to the vertex is 5 cm. We have, a = 5. If the origin is taken at the vertex and the axis of the mirror lies along the positive x-axis, the equation of the parabolic section is y2 = 4 (5) x = 20 x Note that x = 45. Thus y2 = 900 Therefore y = ± 30 Hence AB =2y = 2 × 30 = 60 cm. is a deflection of 3 cm at the centre and the deflected beam is in the shape of a parabola. How far from the centre is the deflection 1 cm? Solution Let the vertex be at the lowest point and the axis vertical. Let the coordinate axis be chosen as shown in Fig 11.34. Fig 11.34 The equation of the parabola takes the form x2 = 4ay. Since it passes through ⎛ 3 ⎞⎛ 3 ⎞ 36 100 ⎜6, ⎟ , we have (6)2 = 4a ⎜⎟ , i.e., a = × = 300 m⎝ 100⎠⎝100 ⎠ 12 12Let AB be the deflection of the beam which is m. Coordinates of B are (x, ).100 100 2Therefore x2 = 4 × 300 × = 24100i.e. x = = 26metres24 Example 19 A rod AB of length 15 cm rests in between two coordinate axes in such a way that the end point A lies on x-axis and end point B lies on y-axis. A point P(x, y) is taken on the rod in such a way that AP = 6 cm. Show that the locus of P is an ellipse. Solution Let AB be the rod making an angle θ with OX as shown in Fig 11.35 and P (x, y) the point on it such that AP = 6 cm. Since AB = 15 cm, we have PB = 9 cm. From P draw PQ and PR perpendiculars on y-axis and x-axis, respectively. Fig 11.35 From Δ PBQ, cos θ = 9 x From Since Δ PRA, sin θ = 6 y cos2 θ + sin2 θ = 1 2 2 9 6 x y⎛⎞ ⎛⎞+⎜⎟ ⎜⎟⎝⎠ ⎝⎠ 1= or 2 2 1 81 36 x y+ = Thus the locus of P is an ellipse. Miscellaneous Exercise on Chapter 11 1. If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus. 2. An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola? 3. The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle. 4. An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end. 5. A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis. 6. Find the area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends of its latus rectum. 7. A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man. 8. An equilateral triangle is inscribed in the parabola y2 = 4 ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle. Summary In this Chapter the following concepts and generalisations are studied. A circle is the set of all points in a plane that are equidistant from a fixed point in the plane. The equation of a circle with centre (h, k) and the radius r is (x – h)2 + (y – k)2 = r2. A parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point in the plane. The equation of the parabola with focus at (a, 0) a > 0 and directrix x = – a is y2 = 4ax. Latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus and whose end points lie on the parabola. Length of the latus rectum of the parabola y2 = 4ax is 4a. An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant. x2 y2 The equation of an ellipse with foci on the x-axis is a2 +b2 =1. Latus rectum of an ellipse is a line segment perpendicular to the major axis through any of the foci and whose end points lie on the ellipse. Length of the latus rectum of the ellipse a2+ b2 =1 is a . The eccentricity of an ellipse is the ratio between the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse. A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant. The equation of a hyperbola with foci on the x-axis is : a2 − b2 =1 Latus rectum of hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose end points lie on the hyperbola. x2 y22b2 x2 y2 22 2xy 2b Length of the latus rectum of the hyperbola : 2 − 2 =1 is : . ab a The eccentricity of a hyperbola is the ratio of the distances from the centre of the hyperbola to one of the foci and to one of the vertices of the hyperbola. Historical Note Geometry is one of the most ancient branches of mathematics. The Greek geometers investigated the properties of many curves that have theoretical and practical importance. Euclid wrote his treatise on geometry around 300 B.C. He was the first who organised the geometric figures based on certain axioms suggested by physical considerations. Geometry as initially studied by the ancient Indians and Greeks, who made essentially no use of the process of algebra. The synthetic approach to the subject of geometry as given by Euclid and in Sulbasutras, etc., was continued for some 1300 years. In the 200 B.C., Apollonius wrote a book called ‘The Conic’ which was all about conic sections with many important discoveries that have remained unsurpassed for eighteen centuries. Modern analytic geometry is called ‘Cartesian’ after the name of Rene Descartes (1596-1650) whose relevant ‘La Geometrie’ was published in 1637. But the fundamental principle and method of analytical geometry were already discovered by Pierre de Fermat (1601-1665). Unfortunately, Fermats treatise on the subject, entitled Ad Locus Planos et So LIDOS Isagoge (Introduction to Plane and Solid Loci) was published only posthumously in 1679. So, Descartes came to be regarded as the unique inventor of the analytical geometry. Isaac Barrow avoided using cartesian method. Newton used method of undetermined coefficients to find equations of curves. He used several types of coordinates including polar and bipolar. Leibnitz used the terms ‘abscissa’, ‘ordinate’ and ‘coordinate’. L’ Hospital (about 1700) wrote an important textbook on analytical geometry. Clairaut (1729) was the first to give the distance formula although in clumsy form. He also gave the intercept form of the linear equation. Cramer (1750) made formal use of the two axes and gave the equation of a circle as ( y – a)2 + (b – x)2 = r He gave the best exposition of the analytical geometry of his time. Monge (1781) gave the modern ‘point-slope’ form of equation of a line as y – y′ = a (x – x′) and the condition of perpendicularity of two lines as aa′ + 1 = 0. S.F. Lacroix (1765–1843) was a prolific textbook writer, but his contributions to analytical geometry are found scattered. He gave the ‘two-point’ form of equation of a line as ββ′ – y– β=(x– α)αα′ – (β–a–b) and the length of the perpendicular from (α, β) on y = ax + b as .1 + a2 ⎛ a–a′⎞ =His formula for finding angle between two lines was tan θ ⎜ ′⎟ . It is, of1 + aa⎝⎠ course, surprising that one has to wait for more than 150 years after the invention of analytical geometry before finding such essential basic formula. In 1818, C. Lame, a civil engineer, gave mE + m′E′ = 0 as the curve passing through the points of intersection of two loci E = 0 and E′ = 0. Many important discoveries, both in Mathematics and Science, have been linked to the conic sections. The Greeks particularly Archimedes (287–212 B.C.) and Apollonius (200 B.C.) studied conic sections for their own beauty. These curves are important tools for present day exploration of outer space and also for research into behaviour of atomic particles. — • —

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