The equation of the ellipse whose axes are th | vedclass

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

Question

(D) Let the equation of the ellipse be $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$.Since it passes through $(-3, 1)$,we have $\frac{9}{a^{2}} + \f

Vedclass · Accepted Answer

$3x^{2} + 5y^{2} - 32 = 0$

Vedclass · Answer

(D) Let the equation of the ellipse be $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$.Since it passes through $(-3, 1)$,we have $\frac{9}{a^{2}} + \frac{1}{b^{2}} = 1$.Given $e^{2} = \frac{2}{5}$,we know $b^{2} = a^{2}(1 - e^{2}) = a^{2}(1 - \frac{2}{5}) = \frac{3}{5}a^{2}$.Substituting $b^{2}$ into the equation: $\frac{9}{a^{2}} + \frac{1}{\frac{3}{5}a^{2}} = 1$.$\frac{9}{a^{2}} + \frac{5}{3a^{2}} = 1$ $\Rightarrow \frac{27 + 5}{3a^{2}} = 1$ $\Rightarrow 3a^{2} = 32$ $\Rightarrow a^{2} = \frac{32}{3}$.Then $b^{2} = \frac{3}{5} 	imes \frac{32}{3} = \frac{32}{5}$.The equation is $\frac{x^{2}}{32/3} + \frac{y^{2}}{32/5} = 1$,which simplifies to $\frac{3x^{2}}{32} + \frac{5y^{2}}{32} = 1$.Thus,$3x^{2} + 5y^{2} = 32$ or $3x^{2} + 5y^{2} - 32 = 0$.

The equation of the ellipse whose axes are the coordinate axes,which passes through the point $(-3, 1)$,and has an eccentricity $e = \sqrt{\frac{2}{5}}$ is:

Similar Questions

An ellipse with its minor and major axes parallel to the coordinate axes passes through $(0,0)$,$(1,0)$,and $(0,2)$. One of its foci lies on the $Y$-axis. The eccentricity of the ellipse is

In an ellipse,the distance between its foci is $6$ and the length of the major axis is $8$. Find its eccentricity.

The equation of an ellipse in its standard form,given the distance between its foci is $2$ units and the length of its latus rectum is $\frac{15}{2}$ units,is

The coordinates of the foci of the ellipse $16x^{2} + 9y^{2} = 144$ are

The distance between the foci of the ellipse $3x^2 + 4y^2 = 48$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module