Question:
Let P(3, 3) be a point on the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\). If the normal to it at P intersects the x-axis at (9, 0) and e is its eccentricity, then the ordered pair \((a^2, e^2)\) is equal to :
Let P(3, 3) be a point on the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\). If the normal to it at P intersects the x-axis at (9, 0) and e is its eccentricity, then the ordered pair \((a^2, e^2)\) is equal to :
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Slope of the normal at P equals the slope of the line joining P and (9, 0); this gives \(b^2=2a^2\), then use the point on the curve.
Updated On: Nov 5, 2026
- \(\left(\frac{9}{2},2\right)\)
- \((9,3)\)
- \(\left(\frac{3}{2},2\right)\)
- \(\left(\frac{9}{2},3\right)\)
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The Correct Option is D
Solution and Explanation
Slope of the tangent at P(3, 3) is \(\frac{b^2x_1}{a^2y_1}=\frac{b^2}{a^2}\), so the slope of the normal is \(-\frac{a^2}{b^2}\).
The normal passes through P(3, 3) and (9, 0), so its slope is \(\frac{0-3}{9-3}=-\frac12\). Hence \(\frac{a^2}{b^2}=\frac12\), i.e. \(b^2=2a^2\).
P lies on the hyperbola: \(\frac{9}{a^2}-\frac{9}{2a^2}=1\Rightarrow \frac{9}{2a^2}=1\Rightarrow a^2=\frac92\), \(b^2=9\).
\(e^2=1+\frac{b^2}{a^2}=1+2=3\).
Answer: (4) \(\left(\frac{9}{2},3\right)\)
The normal passes through P(3, 3) and (9, 0), so its slope is \(\frac{0-3}{9-3}=-\frac12\). Hence \(\frac{a^2}{b^2}=\frac12\), i.e. \(b^2=2a^2\).
P lies on the hyperbola: \(\frac{9}{a^2}-\frac{9}{2a^2}=1\Rightarrow \frac{9}{2a^2}=1\Rightarrow a^2=\frac92\), \(b^2=9\).
\(e^2=1+\frac{b^2}{a^2}=1+2=3\).
Answer: (4) \(\left(\frac{9}{2},3\right)\)
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