The eccentricity of an ellipse centered at th | vedclass

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

Question

(B) The equation of the directrix is $x = 4$,which is parallel to the $y$-axis. This implies the major axis of the ellipse lies along the $x$-axis.Giv

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The equation of the directrix is $x = 4$,which is parallel to the $y$-axis. This implies the major axis of the ellipse lies along the $x$-axis.Given the center is at the origin $(0, 0)$,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The distance of the directrix from the center is given by $\frac{a}{e} = 4$.Given $e = 1/2$,we have $\frac{a}{1/2} = 4$,which implies $a = 2$.Using the relation $b^2 = a^2(1 - e^2)$,we get $b^2 = 2^2(1 - (1/2)^2) = 4(1 - 1/4) = 4(3/4) = 3$.Substituting $a^2 = 4$ and $b^2 = 3$ into the standard equation,we get $\frac{x^2}{4} + \frac{y^2}{3} = 1$.Multiplying by $12$,we obtain $3x^2 + 4y^2 = 12$.

The eccentricity of an ellipse centered at the origin is $1/2$. If one of its directrices is $x = 4$,then the equation of the ellipse is:

Similar Questions

If a circle $(x-1)^2+y^2=r^2$ touches the ellipse $x^2+4y^2=16$ internally,then $r=$

The longest distance of the point $(a, 0)$ from the curve $2x^2+y^2=2x$ is

The distance of the point $\theta$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ from a focus is

The sum of the focal distances of the point $\left(\frac{4}{\sqrt{5}}, \frac{3}{\sqrt{5}}\right)$ on the ellipse $9x^2+4y^2=36$ is

The area (in square units) of the quadrilateral formed by joining the foci of the two ellipses $\frac{x^2}{9}+\frac{y^2}{5}=1$ and $\frac{x^2}{5}+\frac{y^2}{9}=1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module