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Focus of hyperbola is (± 3, 0) and equation of tangent is 2x + y – 4 = 0, find the equation of hyperbola is

1) 4x2 - 5y2 = 20 2) 5x2 - 4y2 = 20 3) 4x2 - 5y2 = 1 4) 5x2 - 4y2 = 1 Solution: (1) 4x2 - 5y2 = 20 Let the hyperbola be x2 / a2 - y2 / b2 =... View Article

The ellipse x² + 4y² = 4 is inscribed in a rectangle aligned with the coordinate axes, which is turn in inscribed in another ellipse that passes through the point (4, 0). Then, the equation of the ellipse is

1) x2 + 12y2 = 16 2) 4x2 + 48y2 = 48 3) 4x2 + 64y2 = 48 4) x2 + 16y2 = 16 Solution: (1) x2 + 12y2 = 16 The given ellipse is x2 / 4 + y2 / 1... View Article

The circle x² + y² – 8x = 0 and hyperbola x²/ 9 – y² / 4 = 1 intersect at the points A and B. Equation of the circle with AB as its diameter is

1) x2 + y2 - 12x + 24 = 0 2) x2 + y2 + 12x + 24 = 0 3) x2 + y2 - 24x - 12 = 0 4) x2 + y2 - 24x - 12 = 0 Solution: (1) x2 + y2 - 12x + 24 = 0... View Article

The circle x² + y² – 8x = 0 and hyperbola x² / 9 – y² / 4 = 1 intersect at the points A and B. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

1) 2x - √5y - 20 = 0 2) 2x - √5y + 4 = 0 3) 3x - 4y + 8 = 0 4) 4x - 3y + 4 = 0 Solution: (2) 2x - √5y + 4 = 0 Equation of tangent to circle... View Article

The parabola y² = 4x and the circle x² + y² – 6x + 1 = 0 will

1) intersect at exactly one point 2) touch each other at two distinct points 3) touch each other at exactly one point 4) intersect at two... View Article

The distance between the vertex of the parabola y = x² – 4x + 3 and the centre of the circle x² = 9 – (y – 3)² is

1) 2√3 units 2) 3√2 units 3) 2√2 units 4) √2 units 5) 2√5 units Solution: (5) 2√5 units y = x2 - 4x + 3 = x2 - 4x + 4 - 1 = (x - 2)2 - 1 y... View Article

Equation of the ellipse whose axes are the axes of coordinates and which passes through the point (-3, 1) and has eccentricity √(2 / 5) is

1) 5x2 + 3y2 - 48 = 0 2) 3x2 + 5y2 - 15 = 0 3) 5x2 + 3y2 - 32 = 0 4) 3x2 + 5y2 - 32 = 0 Solution: (4) 3x2 + 5y2 - 32 = 0 It is given that... View Article

If the eccentricity of the hyperbola x² / a² – y² / b² = 1 is reciprocal to that of the ellipse x² + 4y² = 4. If the hyperbola passes through the focus of the ellipse, then

1) the equation of the hyperbola is x2 / 3 - y2 / 2 = 1 2) a focus of the hyperbola is (2, 0) 3) the eccentricity of the hyperbola is √(5/3)... View Article

Let P and Q are the points on the parabola y² = 4x, if so that the line segment PQ subtends a right angle at the vertex. If PQ intersects the axis of the parabola at R, then the distance of the vertex from R is

1) 1 2) 2 3) 4 4) 6 Solution: (3) 4 (t12, 2t2) and (t22, 2t2) respectively and let the coordinates of R be (x, 0). x = [t12 + t22] / 2, y =... View Article

If the foci of the ellipse x² / 9 + y² = 1 subtend right angle at a point P. Then, the locus of P is

1) x2 + y2 = 1 2) x2 + y2 = 2 3) x2 + y2 = 4 4) x2 + y2 = 8 Solution: (4) x2 + y2 = 8 x2 / 9 + y2 = 1 a = 3, b = 1 e2 = 1 - b2 / a2 = 1 - (1... View Article

The line x = 2y intersects the ellipse x² / 4 + y² = 1 at the points P and Q. The equation of the circle with PQ as diameter is

1) x2 + y2 = 1/2 2) x2 + y2 = 1 3) x2 + y2 = 2 4) x2 + y2 = 5/2 Solution: (4) x2 + y2 = 5/2 x2 / 4 + y2 = 1 4y2 / 4 + y2 = 1 2y2 = 1 y2 = 1... View Article

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

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