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The eccentricity of the ellipse 9x² + 5y² – 30y = 0 is

1) 1/3 2) 2/3 3) 3/4 4) 4/5 Solution: (2) ⅔ 9x2 + 5y2 - 30y = 0 9x2 + 5 [y2 - 6y + 9 - 9] = 0 9x2 + 5 [y - 3]2 = 45 (x2 / 5) + 5 [y - 3]2 /... View Article

The distance between the foci of the hyperbola x² – 3y² – 4x – 6y – 11 = 0 is

1) 4 2) 6 3) 8 4) 10 Solution: (3) 8 x2 - 3y2 - 4x - 6y - 11 = 0 = (x2 - 4x) - 3 (y2 + 2y) = 11 = (x2 - 4x + 4) - 3 (y2 + 2y + 1) = 11 + 1 =... View Article

The equation of the ellipse whose foci are at (± 2, 0) and eccentricity is (1 / 2), is (x² / a²) + (y² / b²) = 1. Then,

1) a2 = 16, b2 = 12 2) a2 = 12, b2 = 16 3) a2 = 16, b2 = 4 4) a2 = 4, b2 = 16 Solution: (1) a2 = 16, b2 = 12 e = 1 / 2 a * (1 / 2) = 2 a = 4... View Article

The sum of the focal distances from any point on the ellipse 9x² + 16y² = 144 is

1) 3 2) 6 3) 8 4) 4 Solution: (3) 8 9x2 + 16y2 = 144 ⇒ (x2 / 16) + (y2 / 9) = 1 a2 = 16, b2 = 9 ⇒ a = 4, b = 3 Length of major axis = 2a =... View Article

Eccentricity of the ellipse x² + 2y² – 2x + 3y + 2 = 0 is

1) 1/√2 2) 1/2 3) 1/2√2 4) 1/√3 Solution: (1) 1/√2 x2 + 2y2 - 2x + 3y + 2 = 0 = (x - 1)2 - 1 + 2 [y2 + (3 / 2)y + 9 / 16 - 9 / 16] + 2 = 0 =... View Article

The vertex of the parabola x² + 2y = 8x – 7 is

1) (9/2, 0) 2) (4, 9/2) 3) (2, 9/2) 4) (4, 7/2) Solution: (2) (4, 9/2) x2 + 2y = 8x - 7 Vertex of the parabola is (0, 0). The equation can... View Article

If the lines joining the foci of the ellipse (x² / a²) + (y² / b²) = 1, where a > b and an extremity of its minor axis are inclined at an angle of 60, then the eccentricity of the ellipse is

1) - (√3/2) 2) 1/2 3) √5/2 4) √7/ 3 5) √3 Solution: (2) 1/2 S (ae, 0) and S’ = (- ae, 0) The coordinates of P are (0, b). ∠SPS’ = 60o OS... View Article

If e₁ and e₂ are the eccentricities of a hyperbola 3x² – 3y² = 25 and its conjugate, then

1) e21 + e22 = 2 2) e21 + e22 = 4 3) e1 + e2 = 4 4) e1 + e2 = √2 Solution: (2) e21 + e22 = 4 Given equation can be rewritten as x2 - y2 = 25... View Article

If e and e₁ are the eccentricities of a hyperbola and its conjugate, then (1 / e²) + (1 / (e₁)²) =

1) 0 2) 1 3) 2 4) 3 Solution: (2) 1 The equation of the hyperbola is (x2 / a2) - (y2 / b2) = 1. b2 = a2 (e2 - 1) e2 - 1 = b2 / a2 e2 = (b2... View Article

The radius of the circle passing through the foci of the ellipse (x² / 16) + (y² / 9) = 1 and having its centre (0, 3) is

1) 4 2) 3/7 3) √12 4) 7/2 Solution: (1) 4 The equation of the ellipse is (x2 / 16) + (y2 / 9) = 1. a = 4 and b = 3 Eccentricity = e = √1 -... View Article

The focal distance of a point on the parabola y² = 16x, whose ordinate is twice the abscissa, is

1) 6 2) 8 3) 10 4) 12 Solution: (2) 8 The given equation is y2 = 16x. Comparing this equation with y2 = 4ax, 4a = 16 a = 16 / 4 = 4 Since... View Article

If the distance between the foci and the distance between the directrices of the hyperbola (x² / a²) – (y² / b²) = 1 are in the ratio 3 : 2, then a : b is

1) √2 : 1 2) √3 : √2 3) 1 : 2 4) 2 : 1 Solution: (1) √2 : 1 Given 2ae / [2a / e] = 3 / 2 ⇒ e2 = 3 / 2 ⇒ (a2 + b2) / a2 = 3 / 2 ⇒ (b / a)2 =... View Article

Let P (a sec θ, b tan θ) and Q (a sec ɸ, b tan ɸ), where θ + ɸ = π / 2, be 2 points on the hyperbola (x² / a²) – (y² / b²) = 1. If (h, k) is the point of intersection of the normals at P and Q, then k =

1) [a2 + b2] / a 2) - [a2 + b2] / a 3) [a2 + b2] / b 4) - [a2 + b2] / b Solution: (4) - [a2 + b2] / b The slope of normal to (x2 / a2) - (y2... View Article

A focus of an ellipse is at the origin. The directrix is the line x = 4 and the eccentricity is 1/2, the length of the semi-major axis is

1) 5/3 2) 8/3 3) 2/3 4) 4/3 Solution: (2) 8/3 Directrix = |(a / e) - ae| = 4 (a / e) - ae = 4 a [(1 / e) - e] = 4 a [2 - (1 / 2)] = 4 a * (3... View Article

A parabola has the origin as its focus and the line x = 2 as the directrix. Then, the vertex of the parabola is at

1) (2, 0) 2) (0, 2) 3) (1, 0) 4) (0, 1) Solution: (3) (1, 0) The vertex is a midpoint of focus and directrix (1, 0) is the vertex.... View Article

If I denotes the semi – latus rectum of the parabola y² = 4ax and SP and SQ denote the segments of any focal chord PQ, S being the focus, then SP, I and SQ are in the relation

1) AP 2) GP 3) HP 4) I2 = SP2 + SQ2 Solution: (3) HP Let 2 points on the parabola be P (at12, 2at1) and Q = (at22, 2at2) SP = √(a - at12)2... View Article

If a hyperbola has a length of its conjugate axis equal to 5 and the distance between its foci is 13, then the eccentricity of the hyperbola is :

1) 9 2) 13 / 12 3) 32/3 4) 64/3 Solution: (2) 13 / 12 2b = 5 2ae = 13 ae = 13 / 2 a2e2 = a2 + b2 a = b e2 = 1 + (b2 / a2) e = 13 / 12... View Article

The equation y² – 8y – x + 19 = 0 represents

1) a parabola whose focus is (1/4, 0) and directrix is x = -1/4 2) a parabola whose vertex is (3, 4) and directrix is x = 11/4 3) a parabola... View Article

Equation of the directrix of parabola 2x² = 14y is equal to

1) y = - (7/4) 2) x = - (7/4) 3) y = (7/4) 4) x= (7/4) Solution: (1) y = - (7/4) The equation of the directrix parabola 2x2 = 14y x2 = 7y a... View Article

In an ellipse, if the lines joining focus to the extremities of the minor axis from an equilateral triangle with the minor axis, then the eccentricity of the ellipse is

1) √3/2 2) √3/4 3) 1/√2 4) √2 / 3 Solution: (1) √3 / 2 Consider the ellipse (x2 / a2) + (y2 / b2) = 1 where a2 > b2 BB’S forms an... View Article