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Conic Section

Multiple Choice Questions

241.

The distance between the directrices of the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$ is

$\frac{9}{\sqrt{5}}$
$\frac{18}{\sqrt{5}}$
$\frac{24}{\sqrt{5}}$
None of these

242.

The equation of the normal to the hyperbola x² - 16y² - 2x - 64y - 72 = 0 at the point (- 4, - 3) is

5x + 16y + 79 = 0
16x + 5y + 97 = 0
16x + 5y + 79 = 0
5x + 16y + 97 = 0

243.

The centre of the circle which circumscribes the square formed by x² - 8x + 12 = 0 and y² - 14y + 45 = 0 is

(4, 5)
(3, 4)
(9, 5)
(4, 7)

244.

If e₁ ande₂ are the eccentricities of the hyperbolas $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 and \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = - 1$ , then the value of $\frac{1}{{e_{1}}^{2}} + \frac{1}{{e_{2}}^{2}}$ is

3
2
1
$\frac{1}{2}$

245.

The point of contact of 3x + 4y + 7 = 0 and x² + y² - 4x - 6y -12 = 0 is

(1, 1)
(- 1, 1)
(1, - 1)
( -1, - 1)

246.
The equation x² - 7xy + 12y² = 0 represents a

circle

pair of parallel straight lines

pair of perpendicular straight lines

pair of non-perpendicular straight lines

pair of non-perpendicular straight lines

$Given equation is x^{2} - 7 xy + 12 y^{2} = 0 On comparing with {ax}^{2} + 2 hxy + {by}^{2} = 0 ∴ a = 1, h = - \frac{7}{2}, b = 12 ∴ m_{1} + m_{2} = - \frac{2 h}{b} and m_{1} m_{2} = \frac{a}{b} \Rightarrow m_{1} + m_{2} = \frac{7}{12} and m_{1} m_{2} = \frac{1}{12} . . . (i) Now, (m_{1} - m_{2}) = \sqrt{{(m_{1} + m_{2})}^{2} - 4 m_{1} m_{2}} = \sqrt{{(\frac{7}{12})}^{2} - 4 \times \frac{1}{12}} = \sqrt{\frac{49 - 48}{144}} = \sqrt{\frac{1}{144}} = \frac{1}{12} \Rightarrow m_{1} - m_{2} = \frac{1}{12} . . . (ii) On solving Eqs . (i) and (ii), we get m_{1} = \frac{1}{3} and m_{2} = \frac{3}{12} Now, \tan (θ) = \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} = \frac{\frac{1}{3} - \frac{3}{12}}{1 + \frac{1}{3} \times \frac{3}{12}} \Rightarrow \tan (θ) = \frac{\frac{1}{12}}{\frac{13}{12}} = \frac{1}{13} \neq 0 Hence, given equation is a non - perpendicular straight lines .$