The normal to the hyperbola x^2a^2-y^29=1 at the point 8,3√3 on it passes through the point

Correct answer: (C) -1,9√3. Given, hyperbola: x^2a^2-y^29=1 it passes through 8,3√3 ∵(64a^2-27)/9=1⇒a^2=16 Now, equation of normal to hyperbola in point form ∵16x/8+9y/(3√3)=16+9 ⇒2x+√3y=25 -1,9√3 satisfies the equation.

Q43 / 90 · JEE Main 2022 (26 Jun Shift 2) · Mathematics

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

A
$(15, - 2 \sqrt{3})$
B
$(9, 2 \sqrt{3})$
Correct answer
C
$(- 1, 9 \sqrt{3})$
D
$(- 1, 6 \sqrt{3})$

Solution

Given,

hyperbola: $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{9} = 1$

it passes through $(8, 3 \sqrt{3})$

$∵ \frac{64}{a^{2}} - \frac{27}{9} = 1 \Rightarrow a^{2} = 16$

Now, equation of normal to hyperbola in point form

$∵ \frac{16 x}{8} + \frac{9 y}{3 \sqrt{3}} = 16 + 9$

$\Rightarrow 2 x + \sqrt{3} y = 25$

$(- 1, 9 \sqrt{3})$ satisfies the equation.

Hence, the correct answer is option: 'C'.

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Solution

Relevant Questions

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