JEE Main 2022 · 26 Jun · Shift 2 · Mathematics Q13

MathematicsSingle correct+4 / -1Question 13 of 90

The normal to the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}9=1$ at the point $\left(8,3\sqrt3\right)$ on it passes through the point :

\left(15,-2\sqrt3\right)

\left(9,2\sqrt3\right)

\left(-1,9\sqrt3\right)

\left(-1,6\sqrt3\right)

Answer and worked solution

Official answer: C

Find $a$ . $\dfrac{64}{a^2}-3=1$ , so $a^2=16$ .

Normal. $\dfrac{a^2x}{x_1}+\dfrac{b^2y}{y_1}=a^2+b^2$ : $2x+\sqrt3y=25$ .

Check. $(-1,9\sqrt3)$ : $-2+27=25$ ✓.

(C)

Topics:Hyperbola Tangents and normals to conics

JEE Main 2022 · 26 Jun · Shift 2, Mathematics question 13