JEE Main 2015 — Mathematics Coordinate Geometry

Equation of the hyperbola

$$\frac{{x}^{2}}{4}-\frac{{y}^{2}}{9}=1$$4x2​−9y2​=1

Focus of hyperbola $({a}_{1}{e}_{1}, 0)$(a1​e1​,0) and $(-{a}_{1}{e}_{1}, 0)$(−a1​e1​,0)

$${a}_{1}=2, {e}_{1}= \sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2}$$a1​=2,e1​=1+49​​=213​​

$\therefore$∴ Foci would be $(+\sqrt{13}, 0)$(+13​,0) and $(-\sqrt{13}, 0)$(−13​,0)

Product of eccentricity would be

$$\frac{\sqrt{13}}{2} \cdot {e}_{2}=\frac{1}{2}$$213​​⋅e2​=21​

$\therefore {e}_{2}= \frac{1}{\sqrt{13}}$∴e2​=13​1​ .

As the major & minor axis of the ellipse coincide with axis of hyperbola then the value of ${a}_{2}$a2​ for ellipse would be $\sqrt{13}$13​ ,

$${e}_{2}=\sqrt{1-\frac{{b}_{2}^{2}}{{a}_{2}^{2}}}$$e2​=1−a22​b22​​​

$$\frac{1}{\sqrt{13}}=\sqrt{1-\frac{{b}_{2}^{2}}{13}}$$13​1​=1−13b22​​​

$${b}_{2}^{2}=12$$b22​=12

$\therefore$∴ Equation of the ellipse would be

$$\frac{{x}^{2}}{13}+\frac{{y}^{2}}{12}=1.$$13x2​+12y2​=1.

Option (1) $\frac{39}{4 \cdot (13)}+\frac{3}{12}=1$4⋅(13)39​+123​=1

Satisfies the equation, hence it lies on the ellipse.

Option (2) $\frac{13}{4 (13)}+\frac{3}{4.12}=1$4(13)13​+4.123​=1

does not lie on the ellipse.

Option (3) $\frac{13}{2(13)}+\frac{6}{12}=1$2(13)13​+126​=1 satisfy.

Option (4) $\frac{13}{13}+0=1$1313​+0=1 satisfy.

So, option $(\frac{\sqrt{13}}{2}, \frac{\sqrt{3}}{2})$(213​​,23​​) is the answer.