JEE Main 2014 · 19 Apr · Shift 1 · Mathematics Q82

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MathematicsSingle correct+4 / -1Question 82 of 90

The tangent at an extremity (in the first quadrant) of latus rectum of the hyperbola $\dfrac{x^2}4-\dfrac{y^2}5=1$ , meets $x$ - axis and $y$ - axis at A and B respectively. Then $(OA)^2-(OB)^2$ , where O is the origin, equals :

-\dfrac{20}9

\dfrac{16}9

-\dfrac43

Answer and worked solution

Official answer: A

$a^2=4$ , $b^2=5$ , $e=\frac32$ . The latus-rectum extremity is $\left(3,\frac52\right)$ . The tangent is $\frac{3x}4-\frac y2=1$ , giving $A=\left(\frac43,0\right)$ and $B=(0,-2)$ .
$OA^2-OB^2=\frac{16}9-4=-\frac{20}9.$

Topics:Hyperbola Tangents and normals to conics

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JEE Main 2014 · 19 Apr · Shift 1, Mathematics question 82