If the ratio of the distance between the foci | vedclass

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

Question

(A) The distance between the foci is $2ae$. The distance between the directrices is $\frac{2a}{e}$. Given the ratio,we have: $\frac{2ae}{2a/e} = \frac

Vedclass · Accepted Answer

$\sqrt{2} : 1$

Vedclass · Answer

(A) The distance between the foci is $2ae$. The distance between the directrices is $\frac{2a}{e}$. Given the ratio,we have: $\frac{2ae}{2a/e} = \frac{3}{2}$ $\Rightarrow e^2 = \frac{3}{2}$. For a hyperbola,the relationship between $a, b,$ and $e$ is $b^2 = a^2(e^2 - 1)$. Therefore,$\frac{b^2}{a^2} = e^2 - 1$. Substituting $e^2 = \frac{3}{2}$: $\frac{b^2}{a^2} = \frac{3}{2} - 1 = \frac{1}{2}$. Thus,$\frac{b}{a} = \frac{1}{\sqrt{2}}$,which implies $a : b = \sqrt{2} : 1$.

If the ratio of the distance between the foci to the distance between the directrices of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $3 : 2$,then $a : b = \dots$

Similar Questions

The straight line $x + y = \sqrt{2}p$ will touch the hyperbola $4x^2 - 9y^2 = 36$,if

The area (in sq units) of the triangle formed by the tangent at $(\sqrt{3}, 0)$ to the hyperbola $x^2-3y^2=3$ with the pair of asymptotes of the hyperbola is

The angle between the asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is equal to:

If the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ has eccentricity $e = \frac{5}{4}$ and the length of the latus rectum equal to $9$,then $ab$ is equal to

The equation $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$ represents:

Vedclass Test Series

Exam Paper Generator

Online Exam Module