Question

Ellipse \( E:\; \dfrac{x^2}{36} + \dfrac{y^2}{16} = 1 \). A hyperbola is confocal with the ellipse and the eccentricity of the hyperbola is equal to \(5\). If the principal axis of the hyperbola is the \(x\)-axis, then the length of the latus rectum of the hyperbola is:

• A. \( \dfrac{96}{\sqrt{5}} \)
• B. \( 24\sqrt{5} \)
• C. \( 18\sqrt{5} \)
• D. \( 12\sqrt{5} \)

Hint:

For confocal conics:
Confocal curves have the same focal distance \(c\)
Hyperbola eccentricity \( e = \frac{c}{a} \)
Latus rectum of hyperbola \( = \frac{2b^2}{a} \)

Answer 1

The Correct Option is A

To find the length of the latus rectum of the hyperbola confocal with the given ellipse, we start with understanding the properties of the ellipse and hyperbola.

The given ellipse is: \(\frac{x^2}{36} + \frac{y^2}{16} = 1\).

For an ellipse of the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the semi-major axis is \(a=6\) and the semi-minor axis is \(b=4\). The foci are given by \(\sqrt{a^2 - b^2}\).

Calculating the focal distance, we have:

\[ c = \sqrt{a^2 - b^2} = \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5}. \]

Since the hyperbola is confocal with this ellipse, it shares the same focal distance \(\pm c = \pm 2\sqrt{5}\).

The equation for a hyperbola with the principal axis along the x-axis is:

\[\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1.\]

The relationship between the hyperbola's focal distance and semi-major axis \(A\) is:

\[ c = \sqrt{A^2 + B^2}. \]

Given that the eccentricity \(e = 5\) for the hyperbola, we can write:

\[ e = \frac{c}{A}. \]

This gives:

\[ 5 = \frac{2\sqrt{5}}{A} \Rightarrow A = \frac{2\sqrt{5}}{5}. \]

Next, to find the latus rectum of the hyperbola, we use the formula for the length of the latus rectum:

\[ \text{Latus rectum} = \frac{2B^2}{A}. \]

From \(B^2 = c^2 - A^2\), we have:

\[ B^2 = (2\sqrt{5})^2 - \left(\frac{2\sqrt{5}}{5}\right)^2, \]\[ B^2 = 20 - \frac{20}{25} = 20 - 0.8 = 19.2. \]

Substitute in the latus rectum formula:

\[ \text{Latus rectum} = \frac{2 \times 19.2}{\frac{2\sqrt{5}}{5}} = \frac{38.4 \times 5}{2\sqrt{5}} = \frac{192}{\sqrt{5}} = \frac{96}{\sqrt{5}}. \]

Thus, the length of the latus rectum of the hyperbola is:

\(\frac{96}{\sqrt{5}}\).