The major and minor axes of an ellipse are al | vedclass

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

Question

(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. The length of the latus rectum is $\frac{2b^2}{a} = 4$,w

Vedclass · Accepted Answer

$x^2+2 y^2=16$

Vedclass · Answer

(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. The length of the latus rectum is $\frac{2b^2}{a} = 4$,which implies $b^2 = 2a$. The distance between the foci is $2ae = 4\sqrt{2}$,so $ae = 2\sqrt{2}$. Squaring both sides,$a^2e^2 = 8$. Since $e^2 = 1 - \frac{b^2}{a^2}$,we have $a^2(1 - \frac{b^2}{a^2}) = 8$,which simplifies to $a^2 - b^2 = 8$. Substituting $b^2 = 2a$,we get $a^2 - 2a - 8 = 0$. Factoring the quadratic equation,$(a - 4)(a + 2) = 0$. Since $a > 0$,we have $a = 4$. Then $b^2 = 2(4) = 8$. The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{8} = 1$,which simplifies to $x^2 + 2y^2 = 16$.

The major and minor axes of an ellipse are along the $X$-axis and $Y$-axis respectively. If its latus rectum is of length $4$ and the distance between the foci is $4 \sqrt{2}$,then the equation of that ellipse is

Similar Questions

If lines $3x + 2y = 10$ and $-3x + 2y = 10$ are tangents at the extremities of the latus rectum of an ellipse whose centre is the origin,then the length of the latus rectum of the ellipse is:

The equations $x = a \cos \theta$ and $y = b \sin \theta$ with $a > b$ represent a conic section. Its eccentricity $e$ is given by:

The equation $\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=4$,where $-2 < x < 2$,represents a

In an ellipse,the distance between its foci is $6$ and the minor axis is $8$. Then its eccentricity is:

The coordinates of a point,in the parametric form,on the ellipse whose foci are $(-1, 0)$ and $(7, 0)$ and eccentricity $e = \frac{1}{2}$,are

Vedclass Test Series

Exam Paper Generator

Online Exam Module