Question

A hyperbola having the transverse axis of length $\sqrt{2}$ has the same foci as that of the ellipse $3x^2+4y^2=12$, then this hyperbola does not pass through which of the following points -

• A. $\left(\frac{\sqrt{3}}{2},\frac{1}{\sqrt{2}}\right)$
• B. $\left(1,\frac{-1}{\sqrt{2}}\right)$
• C. $\left(-\sqrt{\frac{3}{2}},1\right)$
• D. $\left(\frac{1}{\sqrt{2}},0\right)$

Answer 1

The correct option is A

$\left(\frac{\sqrt{3}}{2},\frac{1}{\sqrt{2}}\right)$

Explanation for the correct option:

Step 1. Find the eccentricity of the hyperbola:

Given Ellipse: $3x^2+4y^2=12$

This can also be written as: $\left(\frac{x^2}{4}\right)+\left(\frac{y^2}{3}\right)=1$

$\Rightarrow$${\left(\frac{x}{2}\right)}^2+{\left(\frac{y}{\sqrt{3}}\right)}^2=1$

$\Rightarrow$ ${b_1}^2={a_1}^2(1-{e_1}^2)$

$\Rightarrow$ $3=4(1-{e_1}^2)$

$\Rightarrow$ $3=4-4{e_1}^2$

$\Rightarrow$ $4{e_1}^2=4-3$

$\Rightarrow$ $4{e_1}^2=1$

$\Rightarrow$ ${e_1}^2=\frac{1}{4}$

$\Rightarrow$ $e_1=\frac{1}{2}$

Step 2. Find the focus of hyperbola:

Focus$=\left(\pm a_1{e}_1,0\right)$

$=\left(\pm \left(2\times \frac{1}{2}\right),0\right)$

$=\left(\pm 1,0\right)$

Length of the transverse axis is $2a_2=\sqrt{2}$

$\Rightarrow$ $a_2=\frac{1}{\sqrt{2}}$

Also, $a_2{e}_2=1$

$\begin{array}{rcl}\therefore e_2& =& \frac{\sqrt{2}}{1}\times 1\\ & =& \sqrt{2}\end{array}$

Step 3. Find the equation of hyperbola:

${b_2}^2={a_2}^2({e_2}^2-1)$

$\Rightarrow$${b_2}^2=\frac{1}{2}\times \left(2-1\right)$

$\Rightarrow$${b_2}^2=\frac{1}{2}$

Equation of hyperbola is $x^2-y^2=\frac{1}{2}$

$\therefore \left(\frac{\sqrt{3}}{2},\frac{1}{\sqrt{2}}\right)$ are points from which hyperbola does not passes through.

Hence, the correct option is (A).