Eccentricity of hyperbola whose length of latus rectum is equal to 8 and length of its conjugate axi

English

Gujarati

Hindi

10-2. Parabola, Ellipse, Hyperbola

medium

The eccentricity of the hyperbola whose length of the latus rectum is equal to $8$ and the length of its conjugate axis is equal to half of the distance between its foci is :

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

$\sqrt 3 $

$\frac{4}{3}$

$\frac{4}{{\sqrt 3 }}$

(JEE MAIN-2016)

Solution

We have $\frac{2 b^{2}}{a}=8$

and $2 b=\frac{1}{2}(2 a e)$

$\therefore \frac{2}{a}\left(\frac{a e}{2}\right)^{2}=8$

$a e^{2}=16–(1)$

Now $\frac{2 b^{2}}{a}=8$

$b^{2}=4 a$

$a^{2}\left(e^{2}-1\right)=4 a$

$a e^{2}-a=4$

Substitute the value of $a e^{2}=16$ in eq $(1)$

$16-a=4$

$-a=4-16$

$-a=-12$

$a=12$

$a e^{2}=16$

$e^{2}=\frac{16}{12}$

$e^{2}=\frac{4}{3}$

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

Standard 11

Mathematics

What will be equation of that chord of hyperbola $25{x^2} – 16{y^2} = 400$, whose mid point is $(5, 3)$

hard

Consider a branch of the hyperbola $x^2-2 y^2-2 \sqrt{2} x-4 \sqrt{2} y-6=0$ with vertex at the point $A$. Let $B$ be one of the end points of its latus rectum. If $\mathrm{C}$ is the focus of the hyperbola nearest to the point $\mathrm{A}$, then the area of the triangle $\mathrm{ABC}$ is

normal

(IIT-2008)

$C$ the centre of the hyperbola $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$. The tangents at any point $P$ on this hyperbola meets the straight lines $bx – ay = 0$ and $bx + ay = 0$ in the points $Q$ and $R$ respectively. Then $CQ\;.\;CR = $

hard

Eccentricity of the curve ${x^2} – {y^2} = {a^2}$ is

normal

The difference of the focal distance of any point on the hyperbola $9{x^2} – 16{y^2} = 144$, is

easy