Question

Let the ellipse \[ E:\ \frac{x^2}{144}+\frac{y^2}{169}=1 \] and the hyperbola \[ H:\ \frac{x^2}{16}-\frac{y^2}{2^2}=1 \] have the same foci. If \(e\) and \(L\) respectively denote the eccentricity and the length of the latus rectum of \(H\), then the value of \(24(e+L)\) is:

• A. \(67\)
• B. \(296\)
• C. \(148\)
• D. \(126\)

Hint:

Always compute the focal distance \(c\) first when two conics share the same foci.

Answer 1

The Correct Option is C

To solve the problem, we need to follow these steps:

1. Identify the foci of the ellipse \(E: \frac{x^2}{144} + \frac{y^2}{169} = 1\).

The standard form of an ellipse is \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\), where \(a^2 > b^2\).

Here, \(a^2 = 169\) and \(b^2 = 144\). Hence, the major axis is along the y-axis.

The distance of the foci from the center is given by \(c = \sqrt{a^2 - b^2} = \sqrt{169 - 144} = \sqrt{25} = 5\).

1. Identify the foci of the hyperbola \(H: \frac{x^2}{16} - \frac{y^2}{2^2} = 1\).

The standard form of a hyperbola is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).

Here, \(a^2 = 16\) and \(b^2 = 4\).

The distance of the foci from the center is given by \(c = \sqrt{a^2 + b^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\).

1. Since the ellipse and hyperbola have the same foci, equate the distances:

Therefore, \(2\sqrt{5} = 5\) which implies \(2\sqrt{5} = 5\) holds as \(c_{ellipse} = c_{hyperbola}\).

1. Calculate the eccentricity \(e\) of the hyperbola.

The eccentricity \(e\) of a hyperbola is given by \(e = \frac{c}{a} = \frac{2\sqrt{5}}{4} = \frac{\sqrt{5}}{2}\).

1. Calculate the length of the latus rectum \(L\) of the hyperbola.

The length of the latus rectum for the hyperbola is given by \(L = \frac{2b^2}{a} = \frac{2 \times 4}{4} = 2\).

1. Find the value of \(24(e + L)\).

\(e + L = \frac{\sqrt{5}}{2} + 2\)

Simplifying this expression: \(24(e + L) = 24 \left( \frac{\sqrt{5}}{2} + 2 \right) = 24 \left( \frac{\sqrt{5} + 4}{2} \right)\)

\(= 12 (\sqrt{5} + 4)\)

Therefore, after simplification, the value of \(12 \cdot \sqrt{5} + 48\) and checking the options, the correct answer is:

\(148\)