Question

Let \(\lambda x - 2y = \mu\) be a tangent to the hyperbola \(a^2x^2 - y^2 = b^2\). Then \(\left(\frac{\lambda}{a}\right)^2 - \left(\frac{\mu}{b}\right)^2\) is equal to :

• A. \(-2\)
• B. \(-4\)
• C. \(2\)
• D. \(4\)

Hint:

Use the tangency condition \(c^2 = A^2m^2 - B^2\) for the hyperbola \(\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1\), with \(m = \lambda/2\), \(c = -\mu/2\).

Answer 1

The Correct Option is D

Rewrite the hyperbola as \(\frac{x^2}{b^2/a^2} - \frac{y^2}{b^2} = 1\), so \(A^2 = \frac{b^2}{a^2}\), \(B^2 = b^2\).
The line is \(y = \frac{\lambda}{2}x - \frac{\mu}{2}\), so \(m = \frac{\lambda}{2}\), \(c = -\frac{\mu}{2}\).
Tangency condition for \(\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1\): \(c^2 = A^2m^2 - B^2\).
\(\frac{\mu^2}{4} = \frac{b^2}{a^2}\cdot\frac{\lambda^2}{4} - b^2\).
Multiply by \(\frac{4}{b^2}\): \(\frac{\mu^2}{b^2} = \frac{\lambda^2}{a^2} - 4\).
Therefore \(\left(\frac{\lambda}{a}\right)^2 - \left(\frac{\mu}{b}\right)^2 = 4\).
Answer: (4) 4