AP EAMCET · Maths · Hyperbola
If \(3 \sqrt{2} x-4 y=12\) 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 \(11\)
Answer & Solution
Correct Answer
(B) \(7\)
Step-by-step Solution
Detailed explanation
\(y = \frac{3\sqrt{2}}{4}x - 3 \implies m=\frac{3\sqrt{2}}{4}, c=-3\) \(c^2 = a^2m^2 - b^2 \implies (-3)^2 = a^2(\frac{3\sqrt{2}}{4})^2 - b^2\) \(9 = a^2(\frac{18}{16}) - b^2 \implies 9 = \frac{9}{8}a^2 - b^2 \quad (1)\) \(b^2 = a^2(e^2-1) \implies b^2 = a^2((\frac{5}{4})^2-1)\)…
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