Let H_1: frac{x^2}{a^2}-frac{y^2}{b^2}=1 and | vedclass

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

Question

$\frac{2 b^2}{ a }=15 \sqrt{2}$
$1+\frac{ b ^2}{ a ^2}=\frac{5}{2}$
$a =5 \sqrt{2}$
$b=5 \sqrt{3}$
$\frac{2 A^2}{B}=12 \sqrt{5}$
$2 a \cdot 2 B=1

Vedclass · Accepted Answer

$55$

Vedclass · Answer

$\frac{2 b^2}{ a }=15 \sqrt{2}$
$1+\frac{ b ^2}{ a ^2}=\frac{5}{2}$
$a =5 \sqrt{2}$
$b=5 \sqrt{3}$
$\frac{2 A^2}{B}=12 \sqrt{5}$
$2 a \cdot 2 B=100 \sqrt{10}$
$2.5 \sqrt{2} \cdot 2 B=100 \sqrt{10}$
$B=5 \sqrt{5}$
$A=5 \sqrt{6}$
$e _2^2=1+\frac{ A ^2}{B^2}$
$=1+\frac{150}{125}$
$e _2^2=1+\frac{30}{25}$
$25 e _2^2=55$

Similar Questions

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $5 y^{2}-9 x^{2}=36$

Let the foci of a hyperbola be $(1,14)$ and $(1,-12)$. If it passes through the point $(1,6)$, then the length of its latus-rectum is :

The foci of the hyperbola $9{x^2} - 16{y^2} = 144$ are

Let $P(6,3)$ be a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If the normal at the point $P$ intersects the $x$-axis at $(9,0)$, then the eccentricity of the hyperbola is

Find the equation of the hyperbola with foci $(0,\,\pm 3)$ and vertices $(0,\,\pm \frac {\sqrt {11}}{2})$.