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Conic Sections

Multiple Choice Questions

31.

What is the radius of the circle passing through the point (2, 4) and having centre at the intersection of the lines x - y = 4 and 2x + 3y + 70 = 0 ?

3 units
5 units
3 $\sqrt{3}$ units
5 $\sqrt{2}$

32.

What is the equation of the hyperbola having latus rectum and eccentricity 8 and $\frac{3}{\sqrt{5}}$ respectively.

$\frac{x^{2}}{25} - \frac{y^{2}}{20} = 1$
$\frac{x^{2}}{40} - \frac{y^{2}}{20} = 1$
$\frac{x^{2}}{40} - \frac{y^{2}}{30} = 1$
$\frac{x^{2}}{30} - \frac{y^{2}}{25} = 1$

33.

If the ellipse 9x² + 16y² = 144 intercepts the line 3x + 4y = 12, then what is the length of the chord so formed ?

5 units
6 units
8 units
10 units

34.

What is the eccentricity of rectangular hyperbola ?

$\sqrt{2}$
$\sqrt{3}$
$\sqrt{5}$
$\sqrt{6}$

35.

What is the curve which passes through the point (1, 1) and whose slope is $\frac{2 y}{x}$ ?

Circle
Parabola
Ellipse
Hyperbola

36.

Consider the parabola y = x² + 7x + 2 and the straight liney = 3x - 3.

What are the coordinates of the point on the parabola which is closest to the straight line ?

(0, 2)
( - 2, - 8)
( - 7, 2)
(1, 10)

37.

Consider the parabola y = x² + 7x + 2 and the straight liney = 3x - 3.

What is the shortest distance from the above point on the parabola to the line ?

$\frac{\sqrt{10}}{2}$
$\frac{\sqrt{10}}{5}$
$\frac{1}{\sqrt{10}}$
$\frac{\sqrt{5}}{4}$

38.
If the angle between the lines joining the end points of minor axis of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with one of its foci is $\frac{π}{2}$ , then what is the eccentricity of the ellipse ?

$\frac{1}{2}$

$\frac{1}{\sqrt{2}}$

$\frac{\sqrt{3}}{2}$

$\frac{1}{2 \sqrt{2}}$

$\frac{1}{\sqrt{2}}$

$We have a equation of ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1; here a > b & b^{2} = a^{2} (1 - e^{2}) Here major axis length (AA') = 2 a and minor axis length (BB') = 2 b and foci : S \equiv (ae, 0) & S' \equiv (- ae, 0); e = eccentricity Firstly calculate slope of the line BS and BS' \Rightarrow Slope of BS (m_{1}) = \frac{b - 0}{0 - (- ae)} = \frac{b}{ae} \Rightarrow Slope of BS' (m_{2}) = \frac{b - 0}{0 - ae} = \frac{b}{- ae} Given that, angle between BS and BS' is \frac{π}{2} ∴ we know, \tan (θ) = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}|$

$\tan (90 °) = \infty = \frac{1}{0} \frac{1}{0} = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = \Rightarrow 1 + m_{1} m_{2} = 0 ∴ m_{1} m_{2} = - 1 \Rightarrow (\frac{b}{ae}) (- \frac{b}{ae}) = - 1 \Rightarrow - b^{2} = - a^{2} e^{2} \Rightarrow a^{2} (1 - e^{2}) = a^{2} e^{2} [∵ b^{2} = a^{2} (1 - e^{2})] \Rightarrow 1 - e^{2} = e^{2} \Rightarrow 2 e^{2} = 1 \Rightarrow e = \sqrt{\frac{1}{2}}$