The equation of the ellipse referred to its a | vedclass

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

Question

$2 \mathrm{ae}=4 \sqrt{2}$      .......$(1)$

$\frac{2 b^{2}}{a}=4$      .....$(2)$

$b^{2}=a^{2}\left(1-e^{2}\right

Vedclass · Accepted Answer

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

Vedclass · Answer

$2 \mathrm{ae}=4 \sqrt{2}$      .......$(1)$

$\frac{2 b^{2}}{a}=4$      .....$(2)$

$b^{2}=a^{2}\left(1-e^{2}\right)$      .....$(3)$

From $(1),(2)$ and $( 3)$

$a=4, b=2 \sqrt{2}$

So, $a^{2}=16, b^{2}=8$

Equation of ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{8}=1$

$\Rightarrow x^{2}+2 y^{2}=16$

The equation of the ellipse referred to its axes as the axes of coordinates with latus rectum of length $4$ and distance between foci $4 \sqrt 2$ is-

Similar Questions

Equation of the ellipse with eccentricity $\frac{1}{2}$ and foci at $( \pm 1,\;0)$ is

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is :

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis $(0,\, \pm \sqrt{5})$ ends of minor axis $(±1,\,0)$

If the distance between the foci of an ellipse be equal to its minor axis, then its eccentricity is

The distance between the focii of the ellipse $(3x - 9)^2 + 9y^2 =(\sqrt 2 x + y +1)^2$ is-