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The eccentricity of the conic [(x + 2)² / 7] + (y – 1)² = 14 is

1) √7 / 8 2) √6 / 17 3) √3 / 2 4) √6 / 11 5) √6 / 7 Solution: (5) √6 / 7 a2 = 7 * 14 b2 = 14 e = √1 - (b2 / a2) = √1 - (1 / 7) = √6 / 7... View Article

If P is any point on the ellipse (x² / 36) + (y² / 16) = 1 and S and S’ are the foci, then PS + PS’ =

1) 4 2) 8 3) 10 4) 12 Solution: (4) 12 Given, the ellipse is (x2 / 36) + (y2 / 16) = 1 Here a2 = 36, b2 = 16 Since, a > b, so the sum of... View Article

The equation of the latusrectum of the parabola x² + 4x + 2y = 0, is equal to

1) 2y + 3 = 0 2) 3y = 2 3) 2y = 3 4) 3y + 2 = 0 Solution: (3) 2y = 3 (x + 2) 2 = -2 (y - 2) The equation of latus rectum is y - 2 = - 1 / 2... View Article

The equation of a directrix of the ellipse (x² / 16) + (y² / 25) = 1 is

1) 3y = ± 5 2) y = ± 5 3) 3y = ± 25 4) y = ± 3 Solution: (3) 3y = ± 25 It is given that a = 4, b = 5 and e = √1 - (16 / 25) = 3 / 5 The... View Article

The foci of the conic section 25x² + 16y² – 150x = 175 are

1) (0, ± 3) 2) (0, ± 2) 3) (3, ± 3) 4) (0, ± 1) Solution: (3) (3, ± 3) 25x2 + 16y2 - 150x = 175 25 (x2 - 6x + 9 - 9) + 16y2 = 175 25 (x -... View Article

For the hyperbola, (x² / cos² ɑ) – (y² / sin² ɑ) = 1, which of the following remains constant when ɑ varies?

1) Eccentricity 2) Directrix 3) Abscissae of vertices 4) Abscissae of foci Solution: (4) Abscissae of foci ∵ b2 = a2 (e2 - 1) => sin2α =... View Article

Statement I the curve y = – (x² / 2) + x + 1 is symmetric with respect to the line x = 1. Statement II A parabola is symmetric about its axis

1) Statement I is correct, Statement II is correct; Statement II is a correct explanation for Statement I 2) Statement I is correct, Statement... View Article

Equation of the latusrectum of the ellipse 9x² + 4y² – 18x – 8y – 23 = 0

1) y = ±√5 2) y = -√5 3) y = 1 ± √5 4) x = -1± √5 Solution: (3) y = 1 ± √5 9x2 + 4y2 - 18x - 8y - 23 = 0 = (3x - 3)2 + (2y - 2)2 - 13 - 23 =... View Article

An equilateral triangle is inscribed in the parabola y² = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

1) 8a√3 2) 4a√3 3) 2a√3 4) a√3 Solution: (1) 8a√3 Let the 2 vertices of the triangle be Q and R. The points Q and R will have the same... View Article

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