Question

An ellipse passes through the foci of the hyperbola, $9x^2? 4y^2 = 36$ an and minor axes lie along the transverse and conjugate axes of the hyperbola respectively. If the product of eccentricities of the two conics is $\frac{1}{2},$ then which of the following points does not lie on the ellipse ?

• A. $\left(\sqrt{13}, 0\right)$
• B. $\left(\frac{\sqrt{39}}{2},\sqrt{3}\right)$
• C. $\left(\frac{1}{2},\sqrt{13}, \frac{\sqrt{3}}{2}\right)$
• D. $\left(\sqrt{\frac{13}{2}}, \sqrt{6}\right)$

Answer 1

The Correct Option is C

To solve the given problem, we need to find the equation of the ellipse that passes through the foci of the given hyperbola and then determine which point among the options does not lie on this ellipse.

The equation of the hyperbola is given as:

\[9x^2 - 4y^2 = 36\]

Rewriting the equation in standard form:

\[\frac{x^2}{4} - \frac{y^2}{9} = 1\]

This hyperbola has the transverse axis along the x-axis, and its standard form gives us:

• \(a^2 = 4 \Rightarrow a = 2\)
• \(b^2 = 9 \Rightarrow b = 3\)
• \(c^2 = a^2 + b^2 = 4 + 9 = 13 \Rightarrow c = \sqrt{13}\)

The foci of the hyperbola are \((\pm c, 0) = (\pm \sqrt{13}, 0)\).

According to the problem, the ellipse passes through these foci and has its major axis along the y-axis (transverse axis of the hyperbola). Thus, the standard equation of the ellipse can be represented as:

\[\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\]

The product of eccentricities \(e_1 \times e_2 = \frac{1}{2}\), where:

• For the hyperbola: \(e_1 = \frac{c}{a} = \frac{\sqrt{13}}{2}\)
• For the ellipse: \(e_2 = \sqrt{1 - \frac{b^2}{a^2}}\)

Using the provided product of eccentricities:

\[e_1 \times e_2 = \frac{1}{2} = \frac{\sqrt{13}}{2} \times \sqrt{1 - \frac{b^2}{a^2}}\]

Simplifying, we find:

\[\sqrt{1 - \frac{b^2}{a^2}} = \frac{1}{\sqrt{13}}\]

Therefore,

\[1 - \frac{b^2}{a^2} = \frac{1}{13}\]

 

\[\frac{b^2}{a^2} = \frac{12}{13}\]

Choosing a logical assumption to simplify our calculations, let’s assume \(a^2 = 13\), which is valid as it passes through focus of hyperbola:

\[b^2 = \frac{12}{13} \times 13 = 12\]

Thus, the equation of the ellipse is:

\[\frac{x^2}{12} + \frac{y^2}{13} = 1\]

Now, let's test which points lie inside this ellipse:

1. \(x = \frac{1}{2},\ y = \sqrt{13},\ \frac{\sqrt{3}}{2}\): Substituting these values in the ellipse equation:

Therefore, the point \(\left(\frac{1}{2}, \sqrt{13}, \frac{\sqrt{3}}{2}\right)\) does not lie on the ellipse.