Question:
The tangent at an extremity (in the first quadrant) of latus rectum of the hyperbola \(\frac{x^2}{4}-\frac{y^2}{5}=1\), meets \(x\)-axis and \(y\)-axis at A and B respectively. Then \((OA)^2-(OB)^2\), where O is the origin, equals :
The tangent at an extremity (in the first quadrant) of latus rectum of the hyperbola \(\frac{x^2}{4}-\frac{y^2}{5}=1\), meets \(x\)-axis and \(y\)-axis at A and B respectively. Then \((OA)^2-(OB)^2\), where O is the origin, equals :
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Find the first-quadrant end of the latus rectum \((ae,b^2/a)\), write the tangent there, and read off its intercepts.
Updated On: Nov 11, 2026
- \(-\frac{20}{9}\)
- \(\frac{16}{9}\)
- \(4\)
- \(-\frac{4}{3}\)
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The Correct Option is A
Solution and Explanation
\(a^2=4,\ b^2=5\), \(e^2=1+\frac54=\frac94\), so \(e=\frac32\) and \(ae=3\).
First-quadrant extremity of the latus rectum: \(\left(3,\frac{b^2}{a}\right)=\left(3,\frac52\right)\).
Tangent: \(\frac{3x}{4}-\frac{(5/2)y}{5}=1\), i.e. \(\frac{3x}{4}-\frac y2=1\).
x-intercept: \(x=\frac43\Rightarrow OA=\frac43\). y-intercept: \(y=-2\Rightarrow OB=2\).
\((OA)^2-(OB)^2=\frac{16}{9}-4=-\frac{20}{9}\).
Answer: (1) \(-\frac{20}{9}\)
First-quadrant extremity of the latus rectum: \(\left(3,\frac{b^2}{a}\right)=\left(3,\frac52\right)\).
Tangent: \(\frac{3x}{4}-\frac{(5/2)y}{5}=1\), i.e. \(\frac{3x}{4}-\frac y2=1\).
x-intercept: \(x=\frac43\Rightarrow OA=\frac43\). y-intercept: \(y=-2\Rightarrow OB=2\).
\((OA)^2-(OB)^2=\frac{16}{9}-4=-\frac{20}{9}\).
Answer: (1) \(-\frac{20}{9}\)
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