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Conic Section

Multiple Choice Questions

591.

A circle S cuts three circles

$x^{2} + y^{2} - 4 x - 2 y + 4 = 0 x^{2} + y^{2} - 2 x - 4 y + 1 = 0 and x^{2} + y^{2} + 4 x + 2 y + 1 = 0$

Orthogonally. Then the radius of S is

$\sqrt{\frac{29}{8}}$
$\sqrt{\frac{28}{11}}$
$\sqrt{\frac{29}{7}}$
$\sqrt{\frac{29}{5}}$

592.

The distance between the vertex and the focus of the parabola x² - 2x + 3y - 2 = 0 is

$\frac{4}{5}$
$\frac{3}{4}$
$\frac{1}{2}$
$\frac{5}{6}$

593.

If (x_1, y₁) and (x₂, y₂) are the end points of a focal chord of the parabola y² = 5x, then 4x₁x₂ + y₁y₂, is equal to

25
5
0
$\frac{5}{4}$

594.

The distance between the focii of the ellipse
x = 3cos $(θ)$ , y = 4sin $(θ)$ is

$2 \sqrt{7}$
$7 \sqrt{2}$
$\sqrt{7}$
$3 \sqrt{7}$

595.

The equations of the latus rectum of the ellipse
9x² + 25y² - 36x + 50y - 164 = 0 are

x - 4 = 0, x + 2 = 0
x - 6 = 0, x + 2 = 0
x + 6 = 0, x - 2 = 0
x + 4 = 0, x + 5 = 0

596.

The values of m for which the line y = mx + 2
becomes a tangent to the hyperbola 4x² - 9y² = 36 is

$\pm \frac{2}{3}$
$\pm \frac{2 \sqrt{2}}{3}$
$\pm \frac{8}{9}$
$\pm \frac{4 \sqrt{2}}{3}$

597.

The equation of the common tangent drawn to the curves y = 8x and xy = - 1 is

y = 2x + 1
2y = x + 6
y = x + 2
3y = 8x + 2

598.
The area included between the parabola $y = \frac{x^{2}}{4 a} and the curve y = \frac{8 a^{3}}{(x^{2} + 4 a^{2})} is$

$a^{2} (2 π + \frac{2}{3})$

$a^{2} (2 π - \frac{8}{3})$

$a^{2} (π + \frac{4}{3})$

$a^{2} (2 π - \frac{4}{3})$

$a^{2} (2 π - \frac{4}{3})$

$For point of intersection, equate both equations y = \frac{x^{2}}{4 a} and the curve y = \frac{8 a^{3}}{(x^{2} + 4 a^{2})} \Rightarrow x^{2} (x^{2} + 4 a^{2}) = 4 a (8 a^{3}) \Rightarrow x^{4} + 4 a^{2} x^{2} = 32 a^{4} \Rightarrow x^{4} + 4 a^{2} x^{2} - 32 a^{4} = 0 \Rightarrow x^{2} (x^{2} + 8 a^{2}) - 4 a^{2} (x^{2} + 8 a^{2}) = 0 \Rightarrow (x^{2} + 8 a^{2}) (x^{2} - 4 a^{2}) = 0 \Rightarrow x^{2} - 4 a^{2} = 0 or x^{2} + 8 a^{2} = 0 \Rightarrow x^{2} = 4 a^{2} or x^{2} = - 8 a^{2} ∴ x^{2} = 4 a^{2}$

$\Rightarrow x = \pm 2 a \Rightarrow x = 2 a Hence, we take limits from 0 to 2 a . ∴ Area between 2 graphs = 2 \times \int_{0}^{2 a} [\frac{8 a^{3}}{(x^{2} + 4 a^{2})} - \frac{x^{2}}{4 a}] dx = 2 \times [\int_{0}^{2 a} \frac{8 a^{3}}{(x^{2} + 4 a^{2})} dx - \int_{0}^{2 a} \frac{x^{2}}{4 a} dx] = 2 \times [8 a^{3} \times \int_{0}^{2 a} \frac{1}{(x^{2} + 4 a^{2})} dx - \frac{1}{4 a} \int_{0}^{2 a} x^{2} dx] = 2 \times [8 a^{3} {[\frac{1}{2 a} \tan^{- 1} (\frac{x}{2 a})]}_{0}^{2 a} - \frac{1}{4 a} {[\frac{x^{3}}{3}]}_{0}^{2 a}] = 2 \times \{\frac{8 a^{3}}{2 a} \tan^{- 1} (\frac{2 a}{2 a}) - \tan^{- 1} (\frac{0}{2 a}) - \frac{1}{4 a} [\frac{{(2 a)}^{3}}{3} - \frac{{(0)}^{3}}{3}]\}$

$= 2 \times \{4 a^{2} \tan^{- 1} (1) - \tan^{- 1} (0) - \frac{1}{4 a} (\frac{8 a^{3}}{3} - 0)\} = 2 \times \{4 a^{2} (\frac{π}{4} - 0) - \frac{1}{4 a} (\frac{8 a^{3}}{3})\} = 2 \times \{4 a^{2} \times \frac{π}{4} - \frac{1}{4 a} \times \frac{8 a^{3}}{3}\} = 2 \times \{a^{2} π - \frac{2 a^{2}}{3}\} = 2 \{a^{2} (π - \frac{2}{3})\} = 2 a^{2} (π - \frac{2}{3}) = a^{2} (2 π - \frac{4}{3})$

599.

If a circle with radius 2.5 units passes through the points (2, 3) and (5, 7), then its centre is

(1 5, 2)
(7, 10)
(3, 4)
(3 5, 5)

600.

The circumcentre of the triangle formed by the points (1, 2, 3) (3, - 1, 5), (4, 0, - 3) is

(1, 1, 1)
(2, 2, 2)
(3, 3, 3)
$(\frac{7}{2}, \frac{- 1}{2}, 1)$

Previous Year Papers

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Multiple Choice Questions

598. The area included between the parabola y = x24a and the curve y = 8a3x2 + 4a2 is a22π + 23 a22π - 83 a2π + 43 a22π - 43

598.
The area included between the parabola $y = \frac{x^{2}}{4 a} and the curve y = \frac{8 a^{3}}{(x^{2} + 4 a^{2})} is$

$a^{2} (2 π + \frac{2}{3})$

$a^{2} (2 π - \frac{8}{3})$

$a^{2} (π + \frac{4}{3})$

$a^{2} (2 π - \frac{4}{3})$