The equation of the normal to the curve y = {3} x ≤ft( π/6 + y ) at x = 0, is

Correct Answer is : 2x+3y=02x + {3} y = 02x+3y=0

Correct Answer is : 2x+3y=02x + {3} y = 02x+3y=0

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TS EAMCET 19-July-2022 Shift 2 Solved Paper

Section: Mathematics

Question : 70 of 160

Marks: +1, -0

\sin y = \sqrt{3} x \sin \left( \frac{\pi}{6} + y \right)

x = 0

, is

Solution:

Given, equation of curve

\sin y = \sqrt{3} x \sin \left( \frac{\pi}{6} + y \right)

When

y = 0

\sin 0 = \sqrt{3} x \sin \left( \frac{\pi}{6} + 0 \right)

0 = \sqrt{3} x \times \frac{\sqrt{3}}{2}

\Rightarrow x = 0

Now, differentiate equation of curve w.r.t.

y

, we get

\cos y = \sqrt{3} x \cos \left( \frac{\pi}{6} + y \right) + \sqrt{3} \sin \left( \frac{\pi}{6} + y \right) \frac{dx}{dy}

\Rightarrow -\frac{dx}{dy} = \frac{\sqrt{3} x \cos \left( \frac{\pi}{6} + y \right) - \cos y}{\sqrt{3} \sin \left( \frac{\pi}{6} + y \right)}

Slope of normal at

(0,0)

= -\frac{\cos 0}{\sqrt{3} \sin \frac{\pi}{6}} = -\frac{2}{\sqrt{3}}

Equation of normal is

y - 0 = -\frac{2}{\sqrt{3}} (x - 0)

\Rightarrow 2x + \sqrt{3} y = 0

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