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  • A hyperbola having the transverse axis of length  has the same foci as that of the ellipse <img alt="

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

Option: 1

\left(\frac{1}{\sqrt{2}}, 0\right)


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

The given equation of ellipse can be written as

\mathrm{ \frac{x^2}{4}+\frac{y^2}{3}=1}
Now, \mathrm{ c^2=1-\frac{b^2}{a^2} \Rightarrow c=\sqrt{1-\frac{3}{4}} \Rightarrow c=\frac{1}{2}}

Also, foci \mathrm{=( \pm 1,0)}

Now, the hyperbola has the same foci i.e,\mathrm{( \pm 1,0)}.

Let \mathrm{e^{\prime}} be its eccentricity. Then \mathrm{2 a e^{\prime}=2}

But \mathrm{2 a=\sqrt{2} }(Length of transverse axis)

\begin{aligned} &\mathrm{ \Rightarrow \sqrt{2} e^{\prime}=2 \Rightarrow e^{\prime}=\sqrt{2} }\\ &\mathrm{ \therefore b^2=a^2\left(e^{\prime 2}-1\right)=\frac{1}{2}(2-1)=\frac{1}{2}} \end{aligned}

So, equation of hyperbola is

\mathrm{\frac{x^2}{1 / 2}-\frac{y^2}{1 / 2}=1 \Rightarrow 2 x^2-2 y^2=1 }

From the options, only option (d) does not satisfy the above equation.

Posted by

Deependra Verma

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