Question

Let the eccentricity of the hyperbola ${\displaystyle \frac{x^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=1$ be reciprocal to that of the ellipse $x^2+4y^2=4$. If the hyperbola passes through a focus of the ellipse, then
(a) the equation of hyperbola is ${\displaystyle \frac{x^2}{3}}-{\displaystyle \frac{y^2}{2}}=1$
(b) A focus of hyperbola is $(2,0)$
(c) the eccentricity of hyperbola is $\sqrt{{\displaystyle \frac{5}{3}}}$
(d) the equation of hyperbola is $x^2-3y^2=3$

Answer 1

Hint: We can use formula for eccentricity and focus and hyperbola because in the given question we have a relation between the eccentricity of ellipse and hyperbola. Then we can use the focus of the ellipse to get the final answer.

Complete step-by-step solution:
In given question equation of ellipse is $x^2+4y^2=4$.
On dividing both sides from 4 to write it in standard form.
$\Rightarrow {\displaystyle \frac{x^2+4y^2}{4}}={\displaystyle \frac{4}{4}}$
$\Rightarrow {\displaystyle \frac{x^2}{4}}+{\displaystyle \frac{4y^2}{4}}=1$
$\Rightarrow {\displaystyle \frac{x^2}{4}}+y^2=1$
On comparing with ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$
$\Rightarrow a^2=4,b^2=1$ where a is half of the major axis and b is half of the minor axis.
To calculate eccentricity of ellipse we can use $b^2=a^2(1-e^2)$
We can arrange it as
$\Rightarrow b^2=a^2(1-e^2)$
$\Rightarrow {\displaystyle \frac{b^2}{a^2}}=1-e^2$
$\Rightarrow e^2=1-{\displaystyle \frac{b^2}{a^2}}$
On substitute $a^2=4,b^2=1$
$\Rightarrow e^2=1-{\displaystyle \frac{1}{4}}$
$\Rightarrow e^2={\displaystyle \frac{3}{4}}$
$\Rightarrow e=\pm \sqrt{{\displaystyle \frac{3}{4}}}$
$\Rightarrow e={\displaystyle \frac{\sqrt{3}}{2}}or{\displaystyle \frac{-\sqrt{3}}{2}}$
But eccentricity of the ellipse always lies between 0 to 1.
Hence
$\Rightarrow e={\displaystyle \frac{\sqrt{3}}{2}}$
As given in question eccentricity of a hyperbola is reciprocal of the eccentricity of the ellipse.
Let the equation of hyperbola is ${\displaystyle \frac{x^2}{{a_1}^2}}-{\displaystyle \frac{y^2}{{b_1}^2}}=1$
Hence eccentricity of hyperbola is $e={\displaystyle \frac{2}{\sqrt{3}}}$
To calculate eccentricity of hyperbola we can use ${b_1}^2={a_1}^2(e^2-1)$ where $a_1$ is the transverse axis of hyperbola and $b_1$ is conjungate axis of hyperbola
On substituting $e={\displaystyle \frac{2}{\sqrt{3}}}$
$\Rightarrow {b_1}^2={a_1}^2({\left({\displaystyle \frac{2}{\sqrt{3}}}\right)}^2-1)$
$\Rightarrow {b_1}^2={a_1}^2({\displaystyle \frac{4}{3}}-1)$
$\Rightarrow {b_1}^2={a_1}^2\left({\displaystyle \frac{4-3}{3}}\right)$
$\Rightarrow {b_1}^2={\displaystyle \frac{{a_1}^2}{3}}$………………………………………….(i)
Co-ordinate of focus of ellipse is $(\pm ae,0)$ if $a^2>b^2$.
From equation of ellipse $a=2,e={\displaystyle \frac{\sqrt{3}}{2}}$
$\Rightarrow ae=2\times {\displaystyle \frac{\sqrt{3}}{2}}$
$\Rightarrow ae=\sqrt{3}$
Hence co-ordinate of focus of ellipse is $(\pm \sqrt{3},0)$
As given, equation of hyperbola passes through focus of ellipse. Hence it will satisfy equation of hyperbola ${\displaystyle \frac{x^2}{{a_1}^2}}-{\displaystyle \frac{y^2}{{b_1}^2}}=1$
$\Rightarrow {\displaystyle \frac{{\left(\sqrt{3}\right)}^2}{{a_1}^2}}-{\displaystyle \frac{{\left(0\right)}^2}{{b_1}^2}}=1$
$\Rightarrow {\displaystyle \frac{3}{{a_1}^2}}-0=1$
$\Rightarrow {\displaystyle \frac{3}{{a_1}^2}}=1$
$\Rightarrow {a_1}^2=3$
On substituting ${a_1}^2=3$ in equation (i) ${b_1}^2={\displaystyle \frac{{a_1}^2}{3}}$
$\Rightarrow {b_1}^2={\displaystyle \frac{3}{3}}$
$\Rightarrow {b_1}^2=1$
Hence equation of hyperbola is by substituting value of ${a_1}^2$ and ${b_1}^2$
$\Rightarrow {\displaystyle \frac{x^2}{3}}-{\displaystyle \frac{y^2}{1}}=1$
We can simplify by taking L.C.M
$\Rightarrow {\displaystyle \frac{x^2-3y^2}{3}}=1$
$\Rightarrow x^2-3y^2=3$
In general focus of hyperbola is $(\pm a_{1}e,0)$ if ${a_1}^2>{b_1}^2$.
For hyperbola $e={\displaystyle \frac{2}{\sqrt{3}}},a_1=\sqrt{3}$
$\Rightarrow a_{1}e=\sqrt{3}\times {\displaystyle \frac{2}{\sqrt{3}}}$
$\Rightarrow a_{1}e=2$
Hence focus of hyperbola is $(\pm 2,0)$.
So option b and d is correct.

Note: In general if equation of ellipse is ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$ then eccentricity of ellipse can be calculated from relation $b^2=a^2(1-e^2)$. Co-ordinate of focus of ellipse is $(\pm ae,0)$ if $a^2>b^2$.
If equation of hyperbola is ${\displaystyle \frac{x^2}{{a_1}^2}}-{\displaystyle \frac{y^2}{{b_1}^2}}=1$ then eccentricity of hyperbola can be calculated from relation ${b_1}^2={a_1}^2(e^2-1)$.

In general eccentricity(e) of the conic section defines its shape and it is a non negative real number.
For ellipse,$0<e<1$
For hyperbola, $e>1$
In equations of ellipse and hyperbola if any variable is common then it has to be represented separately to avoid any error arising due to common variable used in the equation