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Conic Section

Multiple Choice Questions

401.

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

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

402.

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

403.

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

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

404.

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

405.
The area included between the parabola $y = \frac{x^{2}}{4 a} and the curve y = \frac{8 a^{3}}{(x^{2} + 4 a^{2})} is$

$a^{2} (2 π + \frac{2}{3})$

$a^{2} (2 π - \frac{8}{3})$

$a^{2} (π + \frac{4}{3})$

$a^{2} (2 π - \frac{4}{3})$

$a^{2} (2 π - \frac{4}{3})$

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

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

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

406.

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)

407.

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

408.

The lines y = 2x + $\sqrt{76}$ and 2y + x = 8 touch the ellipse x² + y²= 1. If the point of $\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$ intersection of these two lines lie on a circle, whose centre coincides with the centre of that ellipse, then the equation of that circle is

$x^{2} + y^{2} = 28$
$x^{2} + y^{2} = 12$
$x^{2} + y^{2} = 12$
$x^{2} + y^{2} = {(4 + \sqrt{8})}^{2}$

409.

If lx + my = 1 is a normal to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , then a²m² - b²l² = ?

$\frac{m^{2}}{l^{2}} {(a^{2} + b^{2})}^{2}$
l² + m²(a² + b²)²
$\frac{l^{2}}{m^{2}} {(a^{2} + b^{2})}^{2}$
l²m²(a² + b²)²

410.

$\frac{x - 1}{3 x + 4} < \frac{x - 3}{3 x - 2} holds, for all x in the internal$

$(\frac{- 4}{3}, \frac{2}{3})$
$(\infty, \frac{- 5}{4})$
$(\frac{3}{3}, \infty)$
$(- \infty, \frac{- 5}{4}) \cup (\frac{3}{4}, - \infty)$

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

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

405.
The area included between the parabola $y = \frac{x^{2}}{4 a} and the curve y = \frac{8 a^{3}}{(x^{2} + 4 a^{2})} is$

$a^{2} (2 π + \frac{2}{3})$

$a^{2} (2 π - \frac{8}{3})$

$a^{2} (π + \frac{4}{3})$

$a^{2} (2 π - \frac{4}{3})$