A hyperbola having the transverse axis oflength {2} has the same foci as that of the ellipse3 x²+4 y²=12, then this hyperbola does notpass through which of the following points? [3 Sep 2020 Shift 1]

Correct Answer is : (32,12)≤ft({3/2}, {1}{{2}} )(23,21)

Correct Answer is : (32,12)≤ft({3/2}, {1}{{2}} )(23,21)

Test Index

Hyperbola

Section: Mathematics

Question : 70 of 98

Marks: +1, -0

\sqrt{2}

3 x^{2}+4 y^{2}=12,

Solution:

Ellipse

:\frac{x^{2}}{4}+\frac{y^{2}}{3}=1

eccentricity

=\sqrt{1-\frac{3}{4}}=\frac{1}{2}

\therefore

foci

=(\pm 1,0)

for hyperbola, given

2 a=\sqrt{2} \Rightarrow a=\frac{1}{\sqrt{2}}

\therefore

hyperbola will be

\frac{x^{2}}{\frac{1}{2}}-\frac{y^{2}}{b^{2}}=1

eccentricity

=\sqrt{1+2 b^{2}}

\therefore

foci

=\left(\pm \sqrt{\frac{1+2 b^{2}}{2}}, 0 \right)

Ellipse and hyperbola have same foci

\Rightarrow \sqrt{\frac{1+2 b^{2}}{2}}=1

\Rightarrow b^{2}=\frac{1}{2}

\therefore

Equation of hyperbola

:\frac{x^{2}}{\frac{1}{2}}-\frac{y^{2}}{\frac{1}{2}}=1

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

Clearly

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

does not lie on it.

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