If the line 5x-2y-6=0 is a tangent to the hyperbola 5x²-ky²=12, then the equation of the normal to this hyperbola at the point ({6}, p)(b [AP EAPCET 18 May 2024 Shift 1]

Correct Answer is : 6x−5y=21{6}x - 5y = 216x−5y=21

Correct Answer is : 6x−5y=21{6}x - 5y = 216x−5y=21

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Hyperbola

Section: Mathematics

Question : 34 of 34

Marks: +1, -0

If the line

5x-2y-6=0

5x^2-ky^2=12

(\sqrt{6}, p)(b<0)

Solution:

We have, equarion of hyperbola

5x^2-ky^2=12

\frac{x^2}{\frac{12}{5} - \frac{y^2}{12}} = 1

Here,

a^2 = \frac{12}{5}

and

b^2 = \frac{12}{K}

Equation of tangent of hyperbola is

5x-2y-6=0

y = \frac{5}{2}x - 3

Here,

m = \frac{5}{2}

and

a^2 m^2 - b^2 = 9

\frac{12}{5} \times \frac{25}{4} - \frac{12}{K} = 9 \Rightarrow 15 - 9 = \frac{12}{K}

K=2

Equation of hyperbola is

5x^2-2y^2=12

\frac{x^2}{\frac{12}{5}} - \frac{y^2}{6} = 1

Since,

(\sqrt{6}, p)

lie on hyperbola

\Rightarrow 5(\sqrt{6})^2 - 2(p^3) = 12

\Rightarrow 30 - 2p^2 = 12

p^2 = 9 \qquad p = 23

\because \quad p < 0

\therefore \quad p = -3

Equation of normal of hyperbola of

(\sqrt{6},-3), 3

\Rightarrow \frac{12x}{5\sqrt{6}} - 2y = \frac{12}{5} + 6

\Rightarrow 2\sqrt{6x} - 10y = 42

\Rightarrow \sqrt{6}x - 5y = 42

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