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S and T are the foci of the ellipse (x² / a²) + (y² / b²) = 1 and B is the end of the minor axis. If △STB is an equilateral triangle, then the eccentricity of the ellipse is

1) 1/4 2) 1/3 3) 1/2 4) √3 / 2 Solution: (3) 1/2 STB is equilateral triangle. tan 60° = (OB / OS) = [b / {(1 / 2) 2ae}] --- becuase d... View Article

An ellipse is drawn by taking a diameter of the circle (x – 1)² + y² = 1 as its semi-minor axis and a diameter of the circle x² + (y – 2)² = 4 is semi – major axis, if the centre of the ellipse at the origin and its axis are the coordinates axes, then the equation of the ellipse is

1) 4x2 + y2 = 4 2) x2 + 4y2 = 8 3) 4x2 + y2 = 8 4) x2 + 4y2 = 16 Solution: (4) x2 + 4y2 = 16 Length of semi-minor axis = 2 The length of... View Article

Let S be the focus of the parabola y² = 8x and PQ be the common chord of the circle x² + y² – 2x – 4y = 0 and the given parabola. The area of ∆PQS is

1) 6 sq units 2) 16 sq units 3) 4 sq units 4) 64 sq units Solution: (3) 4 sq units Each point on y2 = 8x is in the form of (2t2, 4t).... View Article

The ellipse E₁: (x² / 9) + (y² / 4) = 1 is inscribed in a rectangle R whose sides are parallel to the coordinate axes. Another ellipse E₂ passing through the point (0, 4) circumscribes the rectangle R. The eccentricity of the ellipse E₂ is

1) √2/2 2) √3/2 3) 1/2 4) 3/4 Solution: (3) 1/2 Let the ellipse be (x2 / a2) + (y2 / b2) = 1 as it is passing through (0, 4) and (3, 2).... View Article

The angle between the tangents drawn from the point (1, 4) to the parabola y² = 4x is

1) 0 2) π/6 3) π/4 4) π/3 Solution: (4) π/3 The equation of the tangent is y = mx + 1/m. It passes through the point (1, 4). m2 - 4m + 1 = 0... View Article

The foci of the ellipse (x² / 16) + (y² / 9) = 1 and the hyperbola (x² / 144) – (y² / 81) = (1 / 25) coincide. Then the value of b² is

1) 1 2) 5 3) 7 4) 9 Solution: (3) 7 The foci of the ellipse are given. (± 4e, 0) = (± (12 / 5) e’, 0)) 4e = (12 / 5) e’ 4 √[16 - b2] / 16 =... View Article

For hyperbola (x² / cos² ɑ) – (y² / sin² ɑ) = 1 which of the following remains constant with the change in ‘ɑ’

1) abscissae of vertices 2) abscissae of foci 3) eccentricity 4) directrix Solution: (2) abscissae of foci (x2 / cos2 ɑ) - (y2 / sin2 ɑ) = 1... View Article

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

1) x2 + y2 - 6y - 7 = 0 2) x2 + y2 - 6y + 7 = 0 3) x2 + y2 - 6y - 5 = 0 4) x2 + y2 - 6y + 5 = 0 Solution: (1) x2 + y2 - 6y - 7 = 0 The given... View Article

If a ≠ 0 and the line 2bx + 3cy + 4d = 0 passes through the points of intersection of the parabolas y² = 4ax and x² = 4ay, then

1) d2 + (2b - 3c)2 = 0 2) d2 + (3b - 2c)2 = 0 3) d2 + (2b + 3c)2 = 0 4) d2 + (3b + 2c)2 = 0 Solution: (3) d2 + (2b + 3c)2 = 0 Solving the... View Article

Let A and B be two distinct points on the parabola y² = 4x. If the axis of the parabola touches a circle of radius 2 having AB as its diameter, then the slope of the line joining A and B can be

1) - (1/2) 2) 1/2 3) 1 4) None of these Solution: (3) 1 Let the 2 points be A (at12, 2at1) and B (at22, 2at2) a = 1, mid point of AB = [(t12... View Article

If the line y – √3x + 3 = 0 cuts the parabola y² = x + 2 at A and B, then PA. PB is equal to [where, P = (√3, 0)]

1) [4 (2 - √3)] / 3 2) [4 (√3 + 2)] / 3 3) 4√3 / 3 4) [2 (√3 + 2)] / 3 Solution: (2) [4 (√3 + 2)] / 3 Given P = (√3, 0) The equation of line... View Article

The locus of a point that moves such that the difference of its distance from two fixed points is always a constant, is

1) a circle 2) a straight line 3) a hyperbola 4) an ellipse Solution: (4) an ellipse The locus of a point that moves such that the sum of... View Article

The eccentricity of the ellipse 25x² + 16y² – 150x – 175 = 0 is

1) 2/5 2) 2/3 3) 4/5 4) 3/4 5) 3/5 Solution: (5) 3/5 25x2 + 16y2 - 150x - 175 = 0 The given equation can be reduced to [x - 3]2 / 16 + [y2... View Article

The eccentricity of the hyperbola (x² / 16) – (y² / 25) = 1 is

1) 3/4 2) 3/5 3) √41/4 4) √41/5 Solution: (3) √41/4 (x2 / 16) - (y2 / 25) = 1 a2 = 16, b2 = 25 e = √(a2 + b2) / a2 = √(16 + 25) / 16 = √41 /... View Article

The distance between the directrices of the hyperbola x = 8 secθ, y = 8 tanθ is

1) 8√2 2) 16√2 3) 4√2 4) 6√2 Solution: (1) 8√2 Given, x = 8 sec θ and y = 8 tan θ => sec θ = x / 8 and tan θ = y / 8 sec2 θ - tan2θ = 1... View Article

If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is

1) 4√3 2) 4/√3 3) √3 / 4 4) 3 √2 Solution: (4) 3 √2 2ae = 6 and 2a/e = 12 ⇒ ae = 3 and a / e = 6 ⇒ a2 = 18 ⇒ b2 = a2 - a2e2 = 18 - 9 = 9... View Article

An ellipse has OB as semi-minor axis F and F’ its foci and the angle FBF’ is a right angle. Then, the eccentricity of the ellipse is

1) 1/√3 2) 1/4 3) 1/2 4) 1/√2 Solution: (4) 1/√2 Since, F and F’ are foci of an ellipse whose coordinates are (ae, 0) and (- ae, 0)... View Article

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