An ellipse E: frac{x^{2}}{a^{2}}+frac{y^{2}}{ | vedclass

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

Question

Hyp : $\frac{y^{2}}{64}-\frac{x^{2}}{49}=1$

An ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through the vertices of the hyperbola

Vedclass · Accepted Answer

$1552$

Vedclass · Answer

Hyp : $\frac{y^{2}}{64}-\frac{x^{2}}{49}=1$

An ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through the vertices of the hyperbola $H: \frac{x^{2}}{49}-\frac{y^{2}}{64}=-1$.

$S_{0} b^{2}=64$

$e_{H}=\sqrt{1+\frac{a^{2}}{b^{2}}}=\sqrt{1+\frac{49}{64}}$

Ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

$e_{E}=\sqrt{1-\frac{a^{2}}{b^{2}}}=\sqrt{1-\frac{a^{2}}{64}}$

$b=8, \sqrt{\frac{1-a^{2}}{64}} 	imes \frac{\sqrt{113}}{8}=\frac{1}{2} \Rightarrow \sqrt{64-a^{2}} 	imes \sqrt{113}=32$

$\left(64-a^{2}ight)=\frac{32^{2}}{113}$

$\Rightarrow a^{2}=64-\frac{32^{2}}{113}$

$l=\frac{2 a^{2}}{b}=\frac{2}{8}\left(64-\frac{32^{2}}{113}ight)=\frac{1552}{113}$

$113\,l=1552$

Similar Questions

If a hyperbola passes through the point $\mathrm{P}(10,16)$ and it has vertices at $(\pm 6,0),$ then the equation of the normal to it at $P$ is

If $(0,\; \pm 4)$ and $(0,\; \pm 2)$ be the foci and vertices of a hyperbola, then its equation is

A tangent to the hyperbola $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{2} = 1$ meets $x-$ axis at $P$ and $y-$ axis at $Q$. Lines $PR$ and $QR$ are drawn such that $OPRQ$ is a rectangle (where $O$ is the origin). Then $R$ lies on

The equation of the tangents to the conic $3{x^2} - {y^2} = 3$ perpendicular to the line $x + 3y = 2$ is