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Question: The value of m, for which the line y = mx + \(\frac{25\sqrt{3}}{3}\), is a normal to the conic \(\fr...

The value of m, for which the line y = mx + 2533\frac{25\sqrt{3}}{3}, is a normal to the conic x216y29\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1, is

A

3\sqrt{3}

B

–2/3\sqrt{3}

C

-32\frac{\sqrt{3}}{2}

D

1

Answer

–2/3\sqrt{3}

Explanation

Solution

We know that the equation of the normal of the conic x2a2y2b2\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 at point ( a sec θ, b tan θ) is

Ax sec θ + bycot θ = a2 + b2

Or y = ab\frac{- a}{b}sin θ x + a2+b2bcotθ\frac{a^{2} + b^{2}}{b\cot\theta}

Comparing above equation with equation y = mx + 2533\frac{25\sqrt{3}}{3} and taking a = 4, b = 3

We get, a2+b2bcotθ=2533\frac{a^{2} + b^{2}}{b\cot\theta} = \frac{25\sqrt{3}}{3}⇒ tan θ = 3\sqrt{3}⇒ θ = 600

And m = - ab\frac{a}{b}sin θ = 43\frac{- 4}{3}sin 600 = 43x32=23\frac{- 4}{3}x\frac{\sqrt{3}}{2} = \frac{- 2}{\sqrt{3}}.