Let H: frac{x^2}{a^2} - frac{y^2}{b^2} = 1 be | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the distance between foci is $2ae = 6$, so $ae = 3$. The distance between directrices is $\frac{2a}{e} = \frac{8}{3}$, so $\frac{a}{e} = \fr

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) Given the distance between foci is $2ae = 6$, so $ae = 3$. The distance between directrices is $\frac{2a}{e} = \frac{8}{3}$, so $\frac{a}{e} = \frac{4}{3}$. Multiplying these, we get $a^2 = 3 	imes \frac{4}{3} = 4$, so $a = 2$. Then $e^2 = \frac{ae}{a/e} = \frac{3}{4/3} = \frac{9}{4}$, so $e = \frac{3}{2}$. Using $b^2 = a^2(e^2 - 1)$, we get $b^2 = 4(\frac{9}{4} - 1) = 4(\frac{5}{4}) = 5$. The hyperbola equation is $\frac{x^2}{4} - \frac{y^2}{5} = 1$. For the line $x = k$, the points of intersection $A$ and $B$ are $(k, y_1)$ and $(k, -y_1)$, where $\frac{k^2}{4} - \frac{y_1^2}{5} = 1$, so $y_1^2 = 5(\frac{k^2}{4} - 1)$. The area of triangle $AOB$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2y_1) 	imes k = k y_1 = 4\sqrt{15}$. Squaring both sides, $k^2 y_1^2 = 16 	imes 15 = 240$. Substituting $y_1^2$, $k^2 	imes 5(\frac{k^2}{4} - 1) = 240 \implies k^2(\frac{k^2 - 4}{4}) = 48 \implies k^4 - 4k^2 - 192 = 0$. Let $u = k^2$, then $u^2 - 4u - 192 = 0 \implies (u - 16)(u + 12) = 0$. Since $k^2 > 0$, $k^2 = 16$, so $k = 4$. Note: The question asks for $a^2$, which is $4$. However, given the options, there might be a typo in the question statement regarding the line $x=a$. If the line was $x=k$ and the result is $a^2=12$, it implies a different set of parameters. Based on the standard calculation, $a^2=4$.

Similar Questions

If ${m_1}$ and ${m_2}$ are the slopes of the tangents to the hyperbola $\frac{x^2}{25} - \frac{y^2}{16} = 1$ which pass through the point $(6, 2)$,then:

The length of the latus rectum of the curve $xy = 7x + 5y$ is

The equation of the pair of asymptotes of the hyperbola $4x^2 - 9y^2 - 24x - 36y - 36 = 0$ is

The product of the perpendicular distances drawn from any point on the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ to its asymptotes is

Two points $A$ and $B$ have coordinates $(1, 0)$ and $(-1, 0)$ respectively,and $Q$ is a point which satisfies the relation $AQ - BQ = \pm 1$. The locus of $Q$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module