The equation of the normal to the curve siny= 3x sin ⁡( 6+y) at x=0, is

Question

Accepted Answer

Correct answer: A. (a) Given equation of curve is siny= 3x sin( 6+y) … … (i ) at x=0 ⇒ sin ⁡y=0 or y=0 now differentiating eq. (i) w.r.t. x cos ⁡y dy dx= 3 sin ⁡( 6+y)+ 3x cos ⁡( 6+y) dy dx Now slope of tangent at x=0 and y=0 dy dx= 3 sin ⁡ ( 6 ) ⇒ dy dx= 32 Hence, slope of the normal is -1 3= -2 3 at (0 ,0 ) Hence equation of the normal to the curve at (0 ,0 ) is (y-0 )= -2 3 (x-0 ) ⇒2x+ 3y=0

$The equation of the normal to the curve sin y = 3 x sin (\frac{π}{6} + y) at x = 0, is$

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