ઉપવલય કે જેની અક્ષો યામાક્ષોની અક્ષો હોય તથા | vedclass

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

Question

Let the equation of the required ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

As it passses through $(-3,1)$ we get,

$\frac{9}{a^{2}}+

Vedclass · Accepted Answer

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

Vedclass · Answer

Let the equation of the required ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

As it passses through $(-3,1)$ we get,

$\frac{9}{a^{2}}+\frac{1}{b^{2}}=2$

$\Rightarrow 9 b^{2}+a^{2}=a^{2} b^{2}$

$=9 a^{2}\left(1-e^{2}\right)+a^{2}=a^{2} \cdot a^{2}\left(1-e^{2}\right)$

$=9 a^{2}\left(1-\frac{2}{5}\right)+a^{2}=a^{4}\left(1-\frac{2}{5}\right)$

$\Rightarrow a^{2}=\frac{32}{5}$

Now $b^{2}=a^{2}\left(1-e^{2}\right)$

$\Rightarrow b^{2}=\frac{32}{3}\left(1-\frac{2}{5}\right)$

$=\frac{32}{5}$

Hence the equation of required ellipse is

$\frac{x^{2}}{\frac{32}{3}}+\frac{y^{2}}{\frac{32}{5}}=1$

(or) $3 x^{2}+5 y^{2}=32$

Similar Questions

જો $P \equiv (x,\;y)$, ${F_1} \equiv (3,\;0)$, ${F_2} \equiv ( - 3,\;0)$ અને $16{x^2} + 25{y^2} = 400$, તો $ P{F_1} + P{F_2}$ = .. . . .