Question

A hyperbola whose transverse axis is along the major axis of the conic, $\frac{x^2}{3} + \frac{y^2}{4} = 4 $ and has vertices at the foci of this conic. If the eccentricity of the hyperbola is $\frac{3}{2}$, then which of the following points does NOT lie on it ?

• A. (0, 2)
• B. $(\sqrt{5}$, 2 $\sqrt{2} )$
• C. $(\sqrt{10}$, 2 $\sqrt{3})$
• D. $( 5, 2 \sqrt{3} )$

Answer 1

The Correct Option is D

We need to find which of the given points does not lie on the hyperbola whose eccentricity is \(\frac{3}{2}\). The transverse axis of the hyperbola is along the major axis of the conic \(\frac{x^2}{3} + \frac{y^2}{4} = 4\).

First, let's understand the given conic, which is an ellipse:

The equation of the ellipse is \(\frac{x^2}{3} + \frac{y^2}{4} = 4\). We can write it in standard form as:

\(\frac{x^2}{12} + \frac{y^2}{16} = 1\)

Here, \(a^2 = 16\) and \(b^2 = 12\), and it is a vertically oriented ellipse as \(a^2 > b^2\).

The foci of the ellipse are given by the formula:

\(c = \sqrt{a^2 - b^2} = \sqrt{16 - 12} = \sqrt{4} = 2\)

The foci are \((0, \pm c) = (0, \pm 2)\).

The vertices of the hyperbola lie at these foci. Therefore, for the hyperbola, the points \((0, \pm 2)\) serve as the vertices.

The standard form of the hyperbola is:

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

Given that the eccentricity \(e = \frac{3}{2}\), we use the formula for eccentricity, \(e = \frac{\sqrt{a^2 + b^2}}{a}\).

Since \(a = 2\), we calculate:

\(\frac{3}{2} = \frac{\sqrt{2^2 + b^2}}{2}\)

\(3 = \sqrt{4 + b^2}\)

\(9 = 4 + b^2\)

\(b^2 = 5\)

Thus, the equation of the hyperbola is:

\(\frac{y^2}{4} - \frac{x^2}{5} = 1\)

Now, we substitute the given points into this equation to check which does not satisfy it:

1. For \((0, 2)\):
2. For \((\sqrt{5}, 2 \sqrt{2})\):
3. For \((\sqrt{10}, 2 \sqrt{3})\):
4. For \((5, 2 \sqrt{3})\):

Thus, the point \((5, 2 \sqrt{3})\) does not lie on the hyperbola.