Question

If the chord of contact of a point \(P\) w.r.t. the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) passes through a fixed point \((\alpha, \beta)\), then the locus of \(P\) is a straight line. The sum of the squares of the intercepts made by this line on the coordinate axes is

• A. \(\frac{a^4}{\alpha^2} + \frac{b^4}{\beta^2}\)
• B. \(\frac{a^4 + b^4}{\alpha^2 + \beta^2}\)
• C. \(\frac{\alpha^2}{4a^2} + \frac{\beta^2}{4b^2}\)
• D. \(\frac{a^2}{4\alpha^2} + \frac{b^2}{4\beta^2}\)

Hint:

For chord of contact passing through fixed point, write line equation in intercept form to calculate sum of squares of intercepts.

Answer 1

The Correct Option is A

Step 1: Chord of contact equation.
For point \(P(x_1, y_1)\) w.r.t hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), chord of contact: \(\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1\)

Step 2: Fixed point condition.
Chord passes through \((\alpha, \beta)\): \(\frac{\alpha x_1}{a^2} - \frac{\beta y_1}{b^2} = 1\)

Step 3: Locus of P.
This is a linear relation between \(x_1\) and \(y_1\): \(\frac{\alpha}{a^2} x_1 - \frac{\beta}{b^2} y_1 = 1\)

Step 4: Convert to intercept form.
\(\frac{x_1}{a^2/\alpha} - \frac{y_1}{b^2/\beta} = 1\)

Step 5: Sum of squares of intercepts.
Intercepts: \(X = a^2/\alpha, Y = b^2/\beta\)
\(\text{Sum of squares} = X^2 + Y^2 = \frac{a^4}{\alpha^2} + \frac{b^4}{\beta^2}\)

Step 6: Final conclusion.
Hence, the required sum is \[ \boxed{\frac{a^4}{\alpha^2} + \frac{b^4}{\beta^2}} \]