Question

The point of intersection of two tangents drawn to the hyperbola \[ \frac{x^2}{a^2} - \frac{y^2}{4} = 1 \] lie on the circle \[ x^2 + y^2 = 5. \] If these tangents are perpendicular to each other, then \( a \) is:

• A. \( 25 \)
• B. \( 5 \)
• C. \( 9 \)
• D. \( 3 \)

Hint:

For hyperbola problems, use the intersection property and apply perpendicular tangent conditions.

Answer 1

The Correct Option is D

Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{4} = 1$. Let the tangents be drawn at points $P$ and $Q$. Let the point of intersection of the tangents be $(h, k)$. The equation of the chord of contact is $\frac{hx}{a^2} - \frac{ky}{4} = 1$. The equation of the pair of tangents from $(h, k)$ to the hyperbola is $$SS_1 = T^2$$ where $S = \frac{x^2}{a^2} - \frac{y^2}{4} - 1$, $S_1 = \frac{h^2}{a^2} - \frac{k^2}{4} - 1$, and $T = \frac{hx}{a^2} - \frac{ky}{4} - 1$. The tangents are perpendicular if the coefficient of $x^2$ + coefficient of $y^2 = 0$. The equation of the pair of tangents is $$\left(\frac{x^2}{a^2} - \frac{y^2}{4} - 1\right)\left(\frac{h^2}{a^2} - \frac{k^2}{4} - 1\right) = \left(\frac{hx}{a^2} - \frac{ky}{4} - 1\right)^2$$ Expanding and equating the coefficients of $x^2$ and $y^2$, we get $$\frac{1}{a^2}\left(\frac{h^2}{a^2} - \frac{k^2}{4} - 1\right) - \frac{1}{4}\left(\frac{h^2}{a^2} - \frac{k^2}{4} - 1\right) = \frac{h^2}{a^4} - \frac{k^2}{16}$$ $$\left(\frac{1}{a^2} - \frac{1}{4}\right)\left(\frac{h^2}{a^2} - \frac{k^2}{4} - 1\right) = \frac{h^2}{a^4} - \frac{k^2}{16}$$ $$\frac{4-a^2}{4a^2}\left(\frac{h^2}{a^2} - \frac{k^2}{4} - 1\right) = \frac{h^2}{a^4} - \frac{k^2}{16}$$ Since the tangents are perpendicular, we use the director circle. The director circle of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $x^2 + y^2 = a^2 - b^2$. In our case, $b^2 = 4$. The director circle is $x^2 + y^2 = a^2 - 4$. The point of intersection of the tangents lies on the circle $x^2 + y^2 = 5$. Comparing the two equations, we have $$a^2 - 4 = 5$$ $$a^2 = 9$$ $$a = 3$$ Thus, the value of $a$ is 3. Final Answer: The final answer is $\boxed{(4)}$