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

For what values of $t$ does the line $x = at^2$ intersect the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at real points?

Let $x^2+y^2=20$ be the director circle of an ellipse $E$ whose major axis is the $X$-axis and minor axis is the $Y$-axis. If the length of the latus rectum of $E$ is $2$,then the distance between its foci is

Which of the following statements are true and which are false? In each case,give a valid reason for your answer.
$r:$ $A$ circle is a particular case of an ellipse.

The area of the greatest rectangle that can be inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

Let $P$ be a parabola with vertex $(2,3)$ and directrix $2x+y=6$. Let an ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ of eccentricity $e=\frac{1}{\sqrt{2}}$ pass through the focus of the parabola $P$. Then the square of the length of the latus rectum of $E$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module