Question

If \( 3\sqrt{2}x - 4y = 12 \) is a tangent to the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) and \(\frac{5}{4}\) is its eccentricity, then \( a^2 - b^2 = \)

• A. 5
• B. 7
• C. 9
• D. 11

Hint:

For a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), tangent: \( y = mx \pm \sqrt{a^2 m^2 - b^2} \). Eccentricity: \( e^2 = 1 + \frac{b^2}{a^2} \).

Answer 1

The Correct Option is B

Rewrite the tangent \( 3\sqrt{2}x - 4y = 12 \): \[ y = \frac{3\sqrt{2}}{4}x - 3 \] For a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the tangent equation is: \[ y = mx \pm \sqrt{a^2 m^2 - b^2} \] Compare: slope \( m = \frac{3\sqrt{2}}{4} \), constant \( \sqrt{a^2 m^2 - b^2} = 3 \). Thus: \[ a^2 \left( \frac{3\sqrt{2}}{4} \right)^2 - b^2 = 9 \] \[ a^2 \cdot \frac{18}{16} - b^2 = 9 \implies \frac{9}{8}a^2 - b^2 = 9 \implies 9a^2 - 8b^2 = 72 \] Given eccentricity \( e = \frac{5}{4} \): \[ e^2 = 1 + \frac{b^2}{a^2} \implies \frac{25}{16} = 1 + \frac{b^2}{a^2} \implies \frac{b^2}{a^2} = \frac{25}{16} - 1 = \frac{9}{16} \] \[ b^2 = \frac{9}{16}a^2 \] Substitute into \( 9a^2 - 8b^2 = 72 \): \[ 9a^2 - 8 \left( \frac{9}{16}a^2 \right) = 72 \implies 9a^2 - \frac{72}{16}a^2 = 72 \implies 9a^2 - \frac{9}{2}a^2 = 72 \] \[ \frac{18a^2 - 9a^2}{2} = 72 \implies \frac{9a^2}{2} = 72 \implies 9a^2 = 144 \implies a^2 = 16 \] \[ b^2 = \frac{9}{16} \cdot 16 = 9 \] \[ a^2 - b^2 = 16 - 9 = 7 \] Option (2) is correct. Options (1), (3), and (4) do not match.