JEE Main 2017 · 2 Apr · Shift 1 · Mathematics Q81

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MathematicsSingle correct+4 / -1Question 81 of 90

A hyperbola passes through the point $P\left(\sqrt2,\sqrt3\right)$ and has foci at $(\pm2,0)$ . Then the tangent to this hyperbola at P also passes through the point :

\left(2\sqrt2,3\sqrt3\right)

\left(\sqrt3,\sqrt2\right)

\left(-\sqrt2,-\sqrt3\right)

\left(3\sqrt2,2\sqrt3\right)

Answer and worked solution

Official answer: A

$a^2+b^2=4$ and $\dfrac2{a^2}-\dfrac3{b^2}=1$ give $a^2=1$ , $b^2=3$ . The tangent at P is $\sqrt2x-\dfrac{\sqrt3y}3=1$ , and $(2\sqrt2,3\sqrt3)$ satisfies it: $4-3=1$ .

Topics:Hyperbola Tangents and normals to conics

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JEE Main 2017 · 2 Apr · Shift 1, Mathematics question 81