The eccentricity of an ellipse,with its centr | vedclass

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

Question

(B) Given that the directrix is $x = 4$,which is parallel to the $y$-axis,the major axis of the ellipse lies along the $x$-axis.Let the equation of th

Vedclass · Accepted Answer

$3{x^2} + 4{y^2} = 12$

Vedclass · Answer

(B) Given that the directrix is $x = 4$,which is parallel to the $y$-axis,the major axis of the ellipse lies along the $x$-axis.Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a > b$.The eccentricity $e$ is given as $\frac{1}{2}$.Using the relation $e^2 = 1 - \frac{b^2}{a^2}$,we have $\frac{b^2}{a^2} = 1 - e^2 = 1 - (\frac{1}{2})^2 = 1 - \frac{1}{4} = \frac{3}{4}$.The equation of the directrix is $x = \frac{a}{e}$. Given $x = 4$,we have $\frac{a}{e} = 4$.Substituting $e = \frac{1}{2}$,we get $a = 4e = 4 	imes \frac{1}{2} = 2$.Now,$b^2 = \frac{3}{4}a^2 = \frac{3}{4} 	imes (2)^2 = \frac{3}{4} 	imes 4 = 3$.Substituting $a^2 = 4$ and $b^2 = 3$ into the ellipse equation,we get $\frac{x^2}{4} + \frac{y^2}{3} = 1$.Multiplying by $12$,we get $3x^2 + 4y^2 = 12$.

The eccentricity of an ellipse,with its centre at the origin,is $\frac{1}{2}$. If one of the directrices is $x = 4$,then the equation of the ellipse is

Similar Questions

The product of the perpendicular distances drawn from the points $(3,0)$ and $(-3,0)$ to the tangent of the ellipse $\frac{x^2}{36}+\frac{y^2}{27}=1$ at the point $\left(3, \frac{9}{2}\right)$ is:

Find the locus of the midpoint of the portion of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ intercepted between the axes.

If the curves $\frac{x^{2}}{a}+\frac{y^{2}}{b}=1$ and $\frac{x^{2}}{c}+\frac{y^{2}}{d}=1$ intersect each other at an angle of $90^{\circ}$,then which of the following relations is $TRUE$?

How many real tangents can be drawn to the ellipse $5x^2 + 9y^2 = 32$ from the point $(2, 3)$?

The eccentric angles of the extremities of the latus recta of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module