If the eccentricity of the hyperbola x^2 - y^ | vedclass

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

Question

$\frac{{{x^2}}}{5} - \frac{{{y^2}}}{{5{{\cos }^2}\alpha }} = 1$
$e_1^2 = 1 + \frac{{{b^2}}}{{{a^2}}}$ $= $ $1 +$ $\frac{{5{{\cos }^2}\alpha }}{5

Vedclass · Accepted Answer

$\pi /4$

Vedclass · Answer

$\frac{{{x^2}}}{5} - \frac{{{y^2}}}{{5{{\cos }^2}\alpha }} = 1$
$e_1^2 = 1 + \frac{{{b^2}}}{{{a^2}}}$ $= $ $1 +$ $\frac{{5{{\cos }^2}\alpha }}{5}$ $ = 1 + cos^2\alpha $ ;

parallelly eccentricity of the ellipse 
$\frac{{{x^2}}}{{25{{\cos }^2}\alpha }} + \frac{{{y^2}}}{{25}} = 1$ is $e_2^2 = 1 - \frac{{25{{\cos }^2}\alpha }}{{25}}$ $ = sin^2\alpha$ ;

put $e_1 =\sqrt 3  e_2 \Rightarrow$ $e_1^2  = 3e_2^2$ 
$\Rightarrow$ $1 + cos^2\alpha = 3\sin^2\alpha$

$ \Rightarrow$ $2 = 4 \sin^2\alpha$

$ \Rightarrow $ $\sin \alpha = \frac{1}{{\sqrt 2 }}$

If the eccentricity of the hyperbola $x^2 - y^2 \sec^2 \alpha = 5$ is $\sqrt 3 $ times the eccentricity of the ellipse $x^2 \sec^2 \alpha + y^2 = 25, $ then a value of $\alpha$ is :

Similar Questions

The latus rectum of the hyperbola $9{x^2} - 16{y^2} - 18x - 32y - 151 = 0$ is

The eccentricity of the hyperbola $\frac{{\sqrt {1999} }}{3}({x^2} - {y^2}) = 1$ is

Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ be $\frac{5}{4}$. If the equation of the normal at the point $\left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)$ on the hyperbola is $8 \sqrt{5} x +\beta y =\lambda$, then $\lambda-\beta$ is equal to

The radius of the director circle of the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$, is

The equation of a line passing through the centre of a rectangular hyperbola is $x -y -1 = 0$. If one of the asymptotes is $3x -4y -6 = 0$, the equation of other asymptote is