Consider a hyperbola H having its centre at t | vedclass

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

Question

(A) Let the equation of the hyperbola $H$ be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $b^2 = a^2(e^2 - 1)$.Since $C_1$ is centered at the origin

Vedclass · Accepted Answer

$\frac{28}{3}$

Vedclass · Answer

(A) Let the equation of the hyperbola $H$ be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $b^2 = a^2(e^2 - 1)$.Since $C_1$ is centered at the origin and touches the hyperbola at its vertices $(\pm a, 0)$,its radius is $a$. Given the area of $C_1$ is $36 \pi$,we have $\pi a^2 = 36 \pi$,so $a = 6$.$C_2$ is centered at a focus $(ae, 0)$ and touches the hyperbola at its vertex $(a, 0)$. The distance between the focus and the vertex is $|ae - a| = a(e - 1)$. Thus,the radius of $C_2$ is $r = a(e - 1)$.Given the area of $C_2$ is $4 \pi$,we have $\pi r^2 = 4 \pi$,so $r^2 = 4$,which means $r = 2$.Substituting $a = 6$,we get $6(e - 1) = 2$,so $e - 1 = \frac{1}{3}$,which gives $e = \frac{4}{3}$.Now,$b^2 = a^2(e^2 - 1) = 36 \left( (\frac{4}{3})^2 - 1 ight) = 36 \left( \frac{16}{9} - 1 ight) = 36 \left( \frac{7}{9} ight) = 28$.The length of the latus rectum is $\frac{2b^2}{a} = \frac{2 	imes 28}{6} = \frac{28}{3}$.

Similar Questions

If the chord of contact of tangents from a point $P(x_1, y_1)$ to the parabola $y^2 = 4ax$ touches the parabola $x^2 = 4by$,the locus of $P$ is:

The difference of the focal distances of any point on the hyperbola $9x^2 - 16y^2 = 144$ is

Tangent at any point $\theta$ on the curve $x=35 \sec \theta, y=35 \tan \theta$ is

The equation of the line passing through the points $\left(ct_1, \frac{c}{t_1}\right)$ and $\left(ct_2, \frac{c}{t_2}\right)$ is

If the centre,vertex,and focus of a hyperbola are $(0, 0)$,$(4, 0)$,and $(6, 0)$ respectively,then the equation of the hyperbola is

Vedclass Test Series

Exam Paper Generator

Online Exam Module