Question

Let the eccentricity of the hyperbola
\(H : \frac{x²}{a²} - \frac{y²}{b²} = 1\)
be √(5/2) and length of its latus rectum be 6√2, If y = 2x + c is a tangent to the hyperbola H. then the value of c² is equal to

• A. 18
• B. 20
• C. 24
• D. 32

Answer 1

The Correct Option is B

To solve this problem, we start with the given equation of the hyperbola and the information provided:

1. The hyperbola is given by:

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

2. The eccentricity \( e \) of the hyperbola is given as \( \sqrt{5/2} \).

3. The length of the latus rectum is given as \( 6\sqrt{2} \).

Firstly, we know the formula for the eccentricity of a hyperbola:

\( e = \sqrt{1 + \frac{b^2}{a^2}} \)

Given:

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

Equating the expressions for eccentricity:

\( \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{\frac{5}{2}} \)

Squaring both sides:

\( 1 + \frac{b^2}{a^2} = \frac{5}{2} \)

Solving for \( \frac{b^2}{a^2} \):

\( \frac{b^2}{a^2} = \frac{5}{2} - 1 = \frac{3}{2} \)

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

\( \text{Length of latus rectum} = \frac{2b^2}{a} = 6\sqrt{2} \)

Substituting \( b^2 = \frac{3}{2}a^2 \):

\( \frac{2 \times \frac{3}{2}a^2}{a} = 6\sqrt{2} \)

This simplifies to:

\( 3a = 6\sqrt{2} \)

Solving for \( a \):

\( a = 2\sqrt{2} \)

Substituting \( a = 2\sqrt{2} \) back into \( b^2 = \frac{3}{2}a^2 \):

\( b^2 = \frac{3}{2} \times (2\sqrt{2})^2 = 12 \)

So, \( b = \sqrt{12} = 2\sqrt{3} \).

To find the value of \( c^2 \), we know the general equation of the tangent to the hyperbola is:

\( \frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1 \)

Given that \( y = 2x + c \) is a tangent, substitute \( u = 2 \) and rewrite the tangent equation:

\( \frac{x}{a^2} + \frac{4x + 2c}{b^2} = 1 \)

Substituting \( a^2 = 8 \) and \( b^2 = 12 \) and simplifying, the equation becomes quadratic:

Since it is tangent, the discriminant must be zero.

On simplifying and applying conditions, we get:

\( c^2 = 20 \)

Thus, the correct answer is:

20