The eccentricity of an ellipse E with centre | vedclass

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

Question

(D) For ellipse $E$,eccentricity $e = \frac{\sqrt{3}}{2}$ and directrix $x = \frac{a}{e} = \frac{4\sqrt{6}}{3}$.Thus,$a = \frac{4\sqrt{6}}{3} \cdot \f

Vedclass · Accepted Answer

$\frac{8}{7}$

Vedclass · Answer

(D) For ellipse $E$,eccentricity $e = \frac{\sqrt{3}}{2}$ and directrix $x = \frac{a}{e} = \frac{4\sqrt{6}}{3}$.Thus,$a = \frac{4\sqrt{6}}{3} \cdot \frac{\sqrt{3}}{2} = 2\sqrt{2}$.Since $b^2 = a^2(1 - e^2)$,we have $b^2 = 8(1 - 3/4) = 2$,so $b = \sqrt{2}$.For hyperbola $H$,eccentricity $e_H = a = 2\sqrt{2}$.The length of the latus rectum of $H$ is $\frac{2b_H^2}{a_H} = 2b = 2\sqrt{2}$.Also,for hyperbola $H$,$b_H^2 = a_H^2(e_H^2 - 1) = a_H^2(8 - 1) = 7a_H^2$.Substituting this into the latus rectum equation: $\frac{2(7a_H^2)}{a_H} = 2\sqrt{2} \implies 14a_H = 2\sqrt{2} \implies a_H = \frac{\sqrt{2}}{7}$.The distance between the foci of $H$ is $2a_H e_H = 2 \cdot \frac{\sqrt{2}}{7} \cdot 2\sqrt{2} = \frac{8}{7}$.

Similar Questions

The equation of the common tangent with positive slope to the parabola $y^{2}=8 \sqrt{3} x$ and the hyperbola $4 x^{2}-y^{2}=4$ is

Find the equation of one of the common tangents to the parabola $y^2 = 4x$ and the ellipse $\frac{x^2}{4} + \frac{y^2}{3} = 1$.

Let a tangent to the curve $y^2 = 24x$ meet the curve $xy = 2$ at the points $A$ and $B$. Then the midpoints of such line segments $AB$ lie on a parabola with the

The square of the slope of a common tangent drawn to the circle $4x^2 + 4y^2 = 25$ and the ellipse $4x^2 + 9y^2 = 36$ is

If a tangent to the hyperbola $x^2 - \frac{y^2}{3} = 1$ is also a tangent to the parabola $y^2 = 8x$,then the equation of such a tangent with a positive slope is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module