11.1 Overview 11.1.1 Sections of a cone Let l be a fixed vertical line and mbe another line intersecting it at a fixed point V and inclined to it at an angle α (Fig. 11.1). Fig. 11.1 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 cone and extending indefinitely in both directions (Fig. 11.2). 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 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 (Fig.11.3). 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. 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. (b) When α < β < 90o, the section is an ellipse. (c) When β = α; the section is a parabola. (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. Indeed these curves are important tools for present day exploration of outer space and also for research into the behaviour of atomic particles. We take conic sections as plane curves. For this purpose, it is convenient to use equivalent definition that refer only to the plane in which the curve lies, and refer to special points and lines in this plane called foci and directrices. According to this approach, parabola, ellipse and hyperbola are defined in terms of a fixed point (called focus) and fixed line (called directrix) in the plane. If S is the focus and l is the directrix, then the set of all points in the plane whose distance from S bears a constant ratio e called eccentricity to their distance from l is a conic section. As special case of ellipse, we obtain circle for which e = 0 and hence we study it differently. 11.1.2 Circle A circle is the set of all points in a plane which are at a fixed distance from a fixed point in the plane. The fixed point is called the centre of the circle and the distance from centre to any point on the circle is called the radius of the circle. The equation of a circle with radius r having centre (h, k) is given by (x – h)2 + (y – k)2 = r2 The general equation of the circle is given by x2 + y2 + 2gx + 2fy + c = 0, where g, f and c are constants. (a) The centre of this circle is (– g, – f ) (b) The radius of the circle is g2 + f 2 − c The general equation of the circle passing through the origin is given by x2 + y2 + 2gx + 2fy = 0. General equation of second degree i.e., ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represent a circle if (i) the coefficient of x2 equals the coefficient of y2, i.e., a = b ≠ 0 and (ii) the coefficient of xy is zero, i.e., h = 0. The parametric equations of the circle x2 + y2 = r2 are given by x = r cosθ, y = r sinθ where θ is the parameter and the parametric equations of the circle (x – h)2 + (y – k)2 = r2 are given by x – h = r cosθ, y – k = r sinθ or x = h + r cosθ, y = k + r sinθ. Fig. 11.5 Note: The general equation of the circle involves three constants which implies that at least three conditions are required to determine a circle uniquely. 11.1.3 Parabola A parabola is the set of points P whose distances from a fixed point F in the plane are equal to their distances from a fixed line l in the plane. The fixed point F is called focus and the fixed line l the directrix of the parabola. Fig. 11.6 Standard equations of parabola The four possible forms of parabola are shown below in Fig. 11.7 (a) to (d) The 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 (Fig. 11.7). Main facts about the parabola Focal distance of a point Forms of Parabolas y2 = 4ax y2 = – 4ax x2 = 4ay x2 = – 4ay Axis y = 0 y = 0 x = 0 x = 0 Directix x = – a x = a y = – a y = a Vertex (0, 0) (0, 0) (0, 0) (0, 0) Focus (a, 0) (– a, 0) (0, a) (0, – a) Length of latus rectum 4a 4a 4a 4a Equations of latus rectum x = a x = – a y = a y = – a Let the equation of the parabola be y2 = 4ax and P(x, y) be a point on it. Then the distance of P from the focus (a, 0) is called the focal distance of the point, i.e., =| x + a | 11.1.4 Ellipse An ellipse is the set of points in a plane, the sum of whose distances from two fixed points is constant. Alternatively, an ellipse is the set of all points in the plane whose distances from a fixed point in the plane bears a constant ratio, less than, to their distance from a fixed line in the plane. The fixed point is called focus, the fixed line a directrix and the constant ratio (e) the centricity of the ellipse. We have two standard forms of the ellipse, i.e., 22 22xy xy(i) += 1 and (ii) += 1,22 22ab ba In both cases a > b and b2= a2(1 – e2), e < 1. In (i) major axis is along x-axis and minor along y-axis and in (ii) major axis is alongy-axis and minor along x-axis as shown in Fig. 11.8 (a) and (b) respectively. Fig. 11.8 Forms of the ellipse 2 2 2 2 1x y a b + = 2 2 2 2 1x y b a + = a > b a > b Equation of major axis y = 0 x = 0 Length of major axis 2a 2a Equation of Minor axis x = 0 y = 0 Length of Minor axis 2b 2b Directrices x = ± a e y = ± a e Equation of latus rectum x = ± ae y = ± ae Length of latus rectum 2b2 a 2b2 a Centre (0, 0) (0, 0) Focal Distance x2 y2 The focal distance of a point (x, y) on the ellipse a2 + b2 = 1 is a – e | x | from the nearer focus a + e | x | from the farther focus Sum of the focal distances of any point on an ellipse is constant and equal to the length of the major axis. 11.1.5 Hyperbola A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points is constant. Alternatively, a hyperbola is the set of all points in a plane whose distances from a fixed point in the plane bears a constant ratio, greater than 1, to their distances from a fixed line in the plane. The fixed point is called a focus, the fixed line a directrix and the constant ratio denoted by e, the ecentricity of the hyperbola. We have two standard forms of the hyperbola, i.e., 22 22xy yx(i) −=1 and (ii) −= 122 22ab ab Here b2 = a2 (e2 – 1), e > 1. In (i) transverse axis is along x-axis and conjugate axis along y-axis where as in (ii) transverse axis is along y-axis and conjugate axis along x-axis. Fig. 11.9 Main facts about the Hyperbola Forms of the hyperbola 2 2 2 2 1x y a b − = 2 2 2 2 1y x a b − = Equation of transverse axis y = 0 x = 0 Equation of conjugate axis x = 0 y = 0 Length of transverse axis 2a 2a Foci (± ae, 0) (0, ± ae) Equation of latus rectum x = ± ae y = ± ae Length of latus rectum 2b2 a 2b2 a Centre (0, 0) (0, 0) CONIC SECTIONS 193 Focal distance x2 y2 The focal distance of any point (x, y) on the hyperbola 2 − 2 =1 is ab e | x | – a from the nearer focus e | x | + a from the farther focus Differences of the focal distances of any point on a hyperbola is constant and equal to the length of the transverse axis. Parametric equation of conics Conics Parametric equations (i) Parabola : y2 = 4ax x = at2, y = 2at; – ∞ < t < ∞ x2 y2 (ii) Ellipse : a2 + b2 =1 x = a cosθ, y = b sinθ; 0 ≤ θ≤ 2π x2 y2 (iii) Hyperbola : 2 − 2 =1 x = a secθ, y = b tanθ, where ab π ππ 3π−<θ< ; <θ< 222 2 11.2 Solved Examples Short Answer Type Example 1 Find the centre and radius of the circle x2 + y2 – 2x + 4y = 8 Solution we write the given equation in the form (x2 – 2x) + ( y2 + 4y) = 8 Now, completing the squares, we get (x2 – 2x + 1) + ( y2 + 4y + 4) = 8 + 1 + 4 (x – 1)2 + (y + 2)2 = 13 Comparing it with the standard form of the equation of the circle, we see that the centre of the circle is (1, –2) and radius is 13 . Example 2 If the equation of the parabola is x2 = – 8y, find coordinates of the focus, the equation of the directrix and length of latus rectum. Solution The given equation is of the form x2 = – 4ay where a is positive. Therefore, the focus is on y-axis in the negative direction and parabola opens downwards. Comparing the given equation with standard form, we get a = 2. Therefore, the coordinates of the focus are (0, –2) and the the equation of directrix is y = 2 and the length of the latus rectum is 4a, i.e., 8. Example 3 Given the ellipse with equation 9x2 + 25y2 = 225, find the major and minor axes, eccentricity, foci and vertices. Solution We put the equation in standard form by dividing by 225 and get x2 y2 + =125 9 4 This shows that a = 5 and b = 3. Hence 9 = 25(1 – e2), so e = . Since the denominator5 of x2 is larger, the major axis is alongx-axis, minor axis along y-axis, foci are (4, 0) and (– 4, 0) and vertices are (5, 0) and (–5, 0). 36Example 4 Find the equation of the ellipse with foci at (± 5, 0) and x = as one of5the directrices. a 36 5 Solution We have ae = 5, = which give a2 = 36 or a = 6. Therefore, e = 6. e 5 2 25 22xyNow b = a 1 −e =61 −= 11 . Thus, the equation of the ellipse is +=1.36 36 11 Example 5 For the hyperbola 9x2 – 16y2 = 144, find the vertices, foci and eccentricity. x2 y2 Solution The equation of the hyperbola can be written as – =1, so a = 4, b = 316 9 925 5 and 9 = 16 (e2 – 1), so that e2 = 1 , which gives e =+= . Vertices are (± a, 0) =1616 4 (± 4, 0) and foci are (± ae,0) = (± 5, 0). 5Example 6 Find the equation of the hyperbola with vertices at (0, ± 6) and e = .3 Find its foci. Solution Since the vertices are on the y-axes (with origin at the mid-point), the equation y2 x2 is of the form 2 − 2 =1. ab ⎛ 25 −⎞As vertices are (0, ± 6), a = 6, b2 = a2 (e2 – 1) = 36 ⎜ 91⎟= 64 , so the required⎝⎠ y2 x2 equation of the hyperbola is −=1 and the foci are (0, ± ae) = (0, ± 10).36 64 Long Answer Type Example 7 Find the equation of the circle which passes through the points (20, 3), (19, 8) and (2, –9). Find its centre and radius. Solution By substitution of coordinates in the general equation of the circle given by x2 + y2 + 2gx + 2fy + c = 0, we have 40g + 6f + c = –409 38g + 16 f + c = –425 4g – 18 f + c = – 85 From these three equations, we get g = – 7, f = – 3 and c = –111 Hence, the equation of the circle is x2 + y2 – 14x – 6y – 111= 0 or (x – 7)2 + (y – 3)2 =132 Therefore, the centre of the circle is (7, 3) and radius is 13. Example 8 An equilateral triangle is inscribed in the parabola y2 = 4ax whose one vertex is at the vertex of the parabola. Find the length of the side of the triangle. Solution As shown in the figure APQ denotes the equilateral triangle with its equal sides of length l (say). Here AP = l soAR = l cos30° 3 = l 2 l Also, PR = l sin 30° = .2 ⎛ l 3 l ⎞Thus are the coordinates of the point P lying on the⎜ , ⎟⎝⎠22 parabola y2 = 4ax. Fig. 11.10 l2 ⎛ l 3 ⎞ Therefore, =4a ⎜ ⎟ 3.⎠ ⇒ l = 8 a⎝42 Thus, 8 a3 is the required length of the side of the equilateral triangle inscribed in the parabola y2 = 4ax. Example 9 Find the equation of the ellipse which passes through the point (–3, 1) and 2has eccentricity , with x-axis as its major axis and centre at the origin.5 x2 y2 Solution Let a2 + b2 =1 be the equation of the ellipse passing through the point (–3, 1). 91Therefore, we have 2 + 2 =1. ab or 9b2 + a2 = a2 b2 or 9a2 (a2 – e2) + a2 = a2 a2 (1 – e2) (Using b2 = a2 (1 – e2) 32 or a2 = 3 32 ⎛ 2⎞ 32 Again b2 = a2 (1 – e2) = ⎜1 − ⎟= ⎝⎠3 55 Hence, the required equation of the ellipse is x2 y2 + =132 32 35 or 3x2 + 5y2 = 32. Example 10 Find the equation of the hyperbola whose vertices are (± 6, 0) and one of the directrices is x = 4. Solution As the vertices are on the x-axis and their middle point is the origin, the x2 y2 equation is of the type 2 − 2 = 1. ab a Here b2 = a2 (e2 – 1), vertices are (± a, 0) and directrices are given by x = ± . e CONIC SECTIONS 197 a 3 ⎛ 9 −⎞Thus a = 6, = 4 and so e = which gives b2 = 36 ⎜ 1⎠⎟ = 45 e⎝24 x 2 y 2 Consequently, the required equation of the hyperbola is −= 1 36 45 Objective Type Questions Each of the examples from 11 to 16, has four possible options, out of which one is correct. Choose the correct answer from the given four options (M.C.Q.) Example 11 The equation of the circle in the first quadrant touching each coordinate axis at a distance of one unit from the origin is: (A) x2 + y2 – 2x – 2y + 1= 0 (B) x2 + y2 – 2x – 2y – 1 = 0 (C) x2 + y2 – 2x – 2y = 0 (C) x2 + y2 – 2x + 2y – 1 = 0 Solution The correct choice is (A), since the equation can be written as (x – 1)2 + (y – 1)2 = 1 which represents a circle touching both the axes with its centre (1, 1) and radius one unit. Example 12 The equation of the circle having centre (1, –2) and passing through the point of intersection of the lines 3x + y = 14 and 2x + 5y = 18 is (A) x2 + y2 – 2x + 4y – 20 = 0 (B) x2 + y2 – 2x – 4y – 20 = 0 (C) x2 + y2 + 2x – 4y – 20 = 0 (D) x2 + y2 + 2x + 4y – 20 = 0 Solution The correct option is (A). The point of intersection of 3x + y – 14 = 0 and 2x + 5y – 18 = 0 are x = 4, y = 2, i.e., the point (4, 2) Therefore, the radius is = 916 =5+ and hence the equation of the circle is given by (x – 1)2 + (y + 2)2 =25 or x2 + y2 – 2x + 4y – 20 = 0. Example 13 The area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends of its latus rectum is (A) 12 sq. units (B) 16 sq. units (C) 18 sq. units (D) 24 sq. units Solution The correct option is (C). From the figure, OPQ represent the triangle whose area is to be determined. The area of the triangle 11 = PQ × OF = (12 × 3) = 18 Fig. 11.11 22Example 14 The equations of the lines joining the vertex of the parabolay2 = 6x to the points on it which have abscissa 24 are (A) y ± 2x = 0 (B) 2y ± x = 0 (C) x ± 2y = 0 (D) 2x ± y = 0 Solution (B) is the correct choice. Let P and Q be points on the parabola y2 = 6x and OP, OQ be the lines joining the vertex O to the points P and Q whose abscissa are 24. Thus y2 = 6 × 24 = 144 or y = ± 12. Therefore the coordinates of the points P and Q are (24, 12) and (24, –12) respectively. Hence the lines are 12 y =± x ⇒ 2y =± x .24 Fig. 11.12 Example 15 The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which passes through the points (–3, 1) and (2, –2) is (A) 5x2 + 3y2 32 (B) 3x2 + 5y2 = 32 (C) 5x2 – 3y2 = 32 (D) 3x2 + 5y2 + 32 = 0 x2 y2 Solution (B) is the correct choice. Let 2 + 2 =1 be the equation of the ellipse.ab Then according to the given conditions, we have 91 111+= 1 += 22 and 22ab ab 4 32 32 which gives a2 = and b2 = .35 Hence, required equation of ellipse is 3x2 + 5y2 = 32. Example 16 The length of the transverse axis along x-axis with centre at origin of a hyperbola is 7 and it passes through the point (5, –2). The equation of the hyperbola is according to the given condition, the length of transverse axis, i.e., 2a = 7 ⇒ a = . CONIC SECTIONS 199 (A) 4 49 2x 196 51 − 2y 1= (B) 49 4 2x 51 196 − 2y 1= (C) 4 49 2x 51 196 − 2y 1= (D) none of these Solution (C) is the correct choice. Let 2 2 x a − 2 2 y b 1= represent the hyperbola. Then 7 2 Also, the point (5, –2) lies on the hyperbola, so, we have 4 (25) – 4 = 1 which gives 49 b2 196 b2 = . Hence, the equation of the hyperbola is51 4 x2 −51 y2 = 149 196 State whether the statements in Examples 17 and 18 are correct or not. Justify. Example 17 Circle on which the coordinates of any point are (2 + 4 cosθ, –1 + 4 sinθ) where θ is parameter is given by (x – 2)2 + (y + 1)2 = 16. Solution True. From given conditions, we have x = 2 + 4 cosθ ⇒ (x – 2) = 4 cosθ and y = –1 + 4 sinθ ⇒ y + 1 = 4 sinθ. Squaring and adding, we get (x – 2)2 + (y + 1)2 = 16. Example 18 A bar of given length moves with its extremities on two fixed straight lines at right angles. Any point of the bar describes an ellipse. Solution True. Let P (x, y) be any point on the bar such that PA = a and PB = b, clearly from the Fig. 11.13. Fig. 11.13 x = OL = b cosθ and y = PL = a sinθ x2 y2 These give 2 + 2 = 1, which is an ellipse.ba Fill in the blanks in Examples 19 to 23. Example 19 The equation of the circle which passes through the point (4, 5) and has its centre at (2, 2) is ________________ . Solution As the circle is passing through the point (4, 5) and its centre is (2, 2) so its radius is (4 − 2) 2 + (5 − 2) 2 = 13 . Therefore the required answer is (x – 2)2 + (y – 2)2 = 13. Example 20 A circle has radius 3 units and its centre lies on the line y = x – 1. If it passes through the point (7, 3), its equation is ________________ . Solution Let (h, k) be the centre of the circle. Then k = h – 1. Therefore, the equation of the circle is given by (x – h)2 + [y – (h – 1)]2 = 9 ... (1) Given that the circle passes through the point (7, 3) and hence we get (7 – h)2 + (3 – (h – 1))2 =9 or (7 – h)2 + (4 – h)2 =9 or h2 – 11h + 28 = 0 which gives (h – 7) (h – 4) = 0 ⇒ h = 4 or h = 7 Therefore, the required equations of the circles are x2 + y2 – 8x – 6y + 16 = 0 or x2 + y2 – 14x – 12y + 76 = 0 Example 21 If the latus rectum of an ellipse with axis along x-axis and centre at origin is 10, distance between foci = length of minor axis, then the equation of the ellipse is ________________ . 2b2 Solution Given that =10 and 2ae = 2b ⇒ b = ae a Again, we know that b2 = a2 (1 – e2) 1 or 2a2e2 = a2 ⇒ e = (using b = ae)2 Thus a = b 2 CONIC SECTIONS 201 2b2 Again =10 aor b =5 2. Thus we get a = 10 Therefore, the required equation of the ellipse is x2 y2 + =1100 50 Example 22 The equation of the parabola whose focus is the point (2, 3) and directrix is the line x – 4y + 3 = 0 is ________________ . Solution Using the definition of parabola, we have x − 4 y + 3 ( x − 2) 2 +( y – 3) 2 = 17 Squaring, we get 17 (x2 + y2 – 4x – 6y + 13) = x2 + 16y2 + 9 – 8xy – 24y + 6x or 16x2 + y2 + 8xy – 74x – 78y + 212 = 0 x2 y2 Example 23 The eccentricity of the hyperbola 2 − 2 =1 which passes throughab the points (3, 0) and (3 2 , 2) is ________________ . 22 Solution Given that the hyperbola xa2 − by 2 =1 is passing through the points (3, 0) and 2,2) , so we get a2 = 9 and b2 = 4. Again, we know that b2 = a2 (e2 – 1). This gives 4 = 9(e2 – 1) (3 13 or e2 = 9 13 or e =. 3 Short Answer Type 1. Find the equation of the circle which touches the both axes in first quadrant and whose radius is a. 2at a (1 − t2)2. Show that the point (x, y) given by x = 2 and y = 2 lies on a circle1 + t1 +t for all real values of t such that –1 < t< 1 where a is any given real numbers. 3. If a circle passes through the point (0, 0) (a, 0), (0, b) then find the coordinates of its centre. 4. Find the equation of the circle which touches x-axis and whose centre is (1, 2). 5. If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then find the radius of the circle. [Hint: Distance between given parallel lines gives the diameter of the circle.] 6. Find the equation of a circle which touches both the axes and the line 3x – 4y + 8 = 0 and lies in the third quadrant. [Hint: Let a be the radius of the circle, then (– a, – a) will be centre and perpendicular distance from the centre to the given line gives the radius of the circle.] 7. If one end of a diameter of the circle x2 + y2 – 4x – 6y + 11 = 0 is (3, 4), then find the coordinate of the other end of the diameter. 8. Find the equation of the circle having (1, –2) as its centre and passing through 3x + y = 14, 2x + 5y = 18 9. If the line y = 3 x + k touches the circle x2 + y2 = 16, then find the value of k. [Hint: Equate perpendicular distance from the centre of the circle to its radius]. 10. Find the equation of a circle concentric with the circle x2 + y2 – 6x + 12y + 15 = 0 and has double of its area. [Hint: concentric circles have the same centre.] 11. If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity. 12. Given the ellipse with equation 9x2 + 25y2 = 225, find the eccentricity and foci. 513. If the eccentricity of an ellipse is and the distance between its foci is 10, then8 find latus rectum of the ellipse. 2 14. Find the equation of ellipse whose eccentricity is , latus rectum is 5 and the3 centre is (0, 0). x2 y2 15. Find the distance between the directrices of the ellipse += 1.36 20 16. Find the coordinates of a point on the parabola y2 = 8x whose focal distance is 4. 17. Find the length of the line-segment joining the vertex of the parabola y2 = 4ax and a point on the parabola where the line-segment makes an angle θ to the x-axis. 18. If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola. 19. If the line y = mx + 1 is tangent to the parabola y2 = 4x then find the value of m. [Hint: Solving the equation of line and parabola, we obtain a quadratic equation and then apply the tangency condition giving the value of m]. 20. If the distance between the foci of a hyperbola is 16 and its eccentricity is then obtain the equation of the hyperbola. 21. Find the eccentricity of the hyperbola 9y2 – 4x2 = 36. 322. Find the equation of the hyperbola with eccentricity and foci at (± 2, 0).2Long Answer Type 23. If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle. 24. Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line y – 4x + 3 = 0. 25. Find the equation of a circle whose centre is (3, –1) and which cuts off a chord of length 6 units on the line 2x – 5y + 18 = 0. [Hint: To determine the radius of the circle, find the perpendicular distance from the centre to the given line.] 26. Find the equation of a circle of radius 5 which is touching another circle x2 + y2 – 2x – 4y – 20 = 0 at (5, 5). 27. Find the equation of a circle passing through the point (7, 3) having radius 3 units and whose centre lies on the line y = x – 1. 28. Find the equation of each of the following parabolas (a) Directrix x = 0, focus at (6, 0) (b) Vertex at (0, 4), focus at (0, 2) (c) Focus at (–1, –2), directrix x – 2y + 3 = 0 29. Find the equation of the set of all points the sum of whose distances from the points (3, 0) and (9, 0) is 12. 30. Find the equation of the set of all points whose distance from (0, 4) are of23their distance from the line y = 9. 31. Show that the set of all points such that the difference of their distances from (4, 0) and (– 4, 0) is always equal to 2 represent a hyperbola. 32. Find the equation of the hyperbola with 4 (a) Vertices (± 5, 0), foci (± 7, 0) (b) Vertices (0, ± 7), e = (c) Foci (0, ± 3 10 ), passing through (2, 3) Objective Type Questions State Whether the statements in each of the Exercises from 33 to 40 are True or False. Justify 33. The line x + 3y = 0 is a diameter of the circle x2 + y2 + 6x + 2y = 0. 34. The shortest distance from the point (2, –7) to the circle x2 + y2 – 14x – 10y – 151 = 0 is equal to 5. [Hint: The shortest distance is equal to the difference of the radius and the distance between the centre and the given point.] 35. If the line lx + my = 1 is a tangent to the circle x2 + y2 = a2, then the point (l, m) lies on a circle. [Hint: Use that distance from the centre of the circle to the given line is equal to radius of the circle.] 36. The point (1, 2) lies inside the circle x2 + y2 – 2x + 6y + 1 = 0. 37. The line lx + my + n = 0 will touch the parabola y2 = 4ax if ln = am2. x2 y2 38. If P is a point on the ellipse +=1 whose foci are S and S′, then PS + PS′ = 8. 16 25 x2 y2 39. The line 2x + 3y = 12 touches the ellipse +=2 at the point (3, 2).9 4 40. The locus of the point of intersection of lines 3xy 4−− 3k =0 and 3kx + ky – 43 = 0 for different value of k is a hyperbola whose eccentricity is 2. [Hint:Eliminate k between the given equations] Fill in the Blank in Exercises from 41 to 46. 41. The equation of the circle having centre at (3, – 4) and touching the line 5x + 12y – 12 = 0 is ________________ . [Hint: To determine radius find the perpendicular distance from the centre of the circle to the line.] 42. The equation of the circle circumscribing the triangle whose sides are the lines y = x + 2, 3y = 4x, 2y = 3x is ________________ . 43. An ellipse is described by using an endless string which is passed over two pins. If the axes are 6 cm and 4 cm, the length of the string and distance between the pins are ____________. 44. The equation of the ellipse having foci (0, 1), (0, –1) and minor axis of length 1 is ________________ . 45. The equation of the parabola having focus at (–1, –2) and the directrix x – 2y + 3 = 0 is ________________ . 5 46. The equation of the hyperbola with vertices at (0, ± 6) and eccentricity is3______________ and its foci are ________________ . Choose the correct answer out of the given four options (M.C.Q.) in Exercises 47 to 59. 47. The area of the circle centred at (1, 2) and passing through (4, 6) is (A) 5π (B) 10π (C) 25π (D) none of these 48. Equation of a circle which passes through (3, 6) and touches the axes is (A) x2 + y2 + 6x + 6y + 3 = 0 (B) x2 + y2 – 6x – 6y – 9 = 0 (C) x2 + y2 – 6x – 6y + 9 = 0 (D) none of these 49. Equation of the circle with centre on the y-axis and passing through the origin and the point (2, 3) is (A) x2 + y2 + 13y = 0 (B)3x2 + 3y2 + 13x + 3 = 0 (C) 6x2 + 6y2 – 13x = 0 (D) x2 + y2 + 13x + 3 = 0 50. The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is (A) x2 + y2 = 9a2 (B) x2 + y2 = 16a2 (C) x2 + y2 = 4a2 (D) x2 + y2 = a2 [Hint: Centroid of the triangle coincides with the centre of the circle and the 2 radius of the circle is 3 of the length of the median] 51. If the focus of a parabola is (0, –3) and its directrix is y = 3, then its equation is (A) x2 = –12y (B) x2 = 12y (C) y2 = –12x (D) y2 = 12x 52. If the parabola y2 = 4ax passes through the point (3, 2), then the length of its latus rectum is 24 1 (A) (B) (C) (D) 4 33 3 53. If the vertex of the parabola is the point (–3, 0) and the directrix is the line x + 5 = 0, then its equation is (A) y2 = 8 (x + 3) (B) x2 = 8 (y + 3) (C) y2 = – 8 (x + 3) (D) y2 = 8 ( x + 5) 54. The equation of the ellipse whose focus is (1, –1), the directrix the line x – y – 3 1= 0 and eccentricity is2(A) 7x2 + 2xy + 7y2 – 10x + 10y + 7 = 0 (B) 7x2 + 2xy + 7y2 + 7 = 0 (C) 7x2 + 2xy + 7y2 + 10x – 10y – 7 = 0 (D) none 55. The length of the latus rectum of the ellipse 3x2 + y2 = 12 is 4 (A) 4 (B)3 (C)8 (D) x2 y2 56. If e is the eccentricity of the ellipse 2 + 2 =1 (a < b), thenab (A) b2 = a2 (1 – e2) (B) a2 = b2 (1 – e2) (C) a2 = b2 (e2 – 1) (D) b2 = a2 (e2 – 1) 57. The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half of the distance between the foci is 44 2 (A) (B) (C) (D) none of these3 33 58. The distance between the foci of a hyperbola is 16 and its eccentricity is 2 . Its equation is x 2 y 2 (A) x2 – y2 = 32 (B) −= 1 (C) 2x – 3y2 = 7 (D) none of these 4 9 3 59. Equation of the hyperbola with eccentricty and foci at (± 2, 0) is22 222 22x y 4 xy 4 xy(A) −= (B) −= (C) −= 1 (D) none of these 4 59 999 49