Eccentricity of the hyperbola conjugate to th | vedclass

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

Question

$e_1^2 = 1 + \frac{{{b^2}}}{{{a^2}}}$ $=$ $1 +$$\frac{{12}}{4}$ $= 4$ $ \Rightarrow$ $ e_1 = 2$ ;
now $\frac{1}{{e_1^2}} + \frac{1}{{e_2^2}}$ $=

Vedclass · Accepted Answer

$\frac{2}{{\sqrt 3 }}$

Vedclass · Answer

$e_1^2 = 1 + \frac{{{b^2}}}{{{a^2}}}$ $=$ $1 +$$\frac{{12}}{4}$ $= 4$ $ \Rightarrow$ $ e_1 = 2$ ;
now $\frac{1}{{e_1^2}} + \frac{1}{{e_2^2}}$ $= 1$ 
$\frac{1}{{e_2^2}}$ $=$ $ 1 -$ $\frac{1}{4}$ $ =$ $\frac{3}{4}$
$ \Rightarrow$ $e_2^2$  $=$ $\frac{4}{3}$  $\Rightarrow$  $e_2 $ $= $ $\frac{2}{{\sqrt 3 }}$

Eccentricity of the hyperbola conjugate to the hyperbola $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{{12}} = 1$ is

Similar Questions

The graph of the conic $ x^2 - (y - 1)^2 = 1$ has one tangent line with positive slope that passes through the origin. the point of tangency being $(a, b). $ Then Length of the latus rectum of the conic is

If the straight line $x\cos \alpha + y\sin \alpha = p$ be a tangent to the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$, then

Find the equation of the hyperbola satisfying the give conditions: Vertices $(0,\,\pm 5),$ foci $(0,\,±8)$

The equation of the hyperbola whose foci are the foci of the ellipse $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1$ and the eccentricity is $2$, is

The equation of the tangent to the hyperbola $2{x^2} - 3{y^2} = 6$ which is parallel to the line $y = 3x + 4$, is