Question:
Let \(a\) and \(b\) respectively be the semi-transverse and semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation \(9e^2-18e+5=0\). If \(S(5, 0)\) is a focus and \(5x=9\) is the corresponding directrix of this hyperbola, then \(a^2-b^2\) is equal to :
Let \(a\) and \(b\) respectively be the semi-transverse and semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation \(9e^2-18e+5=0\). If \(S(5, 0)\) is a focus and \(5x=9\) is the corresponding directrix of this hyperbola, then \(a^2-b^2\) is equal to :
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Solve the quadratic for \(e\) (take \(e\gt1\)), then use focus \(=(ae,0)\) and \(b^2=a^2(e^2-1)\).
Updated On: Nov 11, 2026
- \(7\)
- \(-7\)
- \(5\)
- \(-5\)
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The Correct Option is B
Solution and Explanation
\(9e^2-18e+5=0\Rightarrow e=\frac{18\pm12}{18}=\frac53\) or \(\frac13\). For a hyperbola \(e\gt1\), so \(e=\frac53\).
Focus \((ae,0)=(5,0)\Rightarrow a\cdot\frac53=5\Rightarrow a=3\). (Directrix \(x=\frac ae=\frac95\), i.e. \(5x=9\), consistent.)
\(b^2=a^2(e^2-1)=9\left(\frac{25}{9}-1\right)=16\).
\(a^2-b^2=9-16=-7\).
Answer: (2) \(-7\)
Focus \((ae,0)=(5,0)\Rightarrow a\cdot\frac53=5\Rightarrow a=3\). (Directrix \(x=\frac ae=\frac95\), i.e. \(5x=9\), consistent.)
\(b^2=a^2(e^2-1)=9\left(\frac{25}{9}-1\right)=16\).
\(a^2-b^2=9-16=-7\).
Answer: (2) \(-7\)
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