# Find the equation of the normal to the curve y = $$2x^2 + 3 sin x$$ at x = 0.

## Solution :

The equation of the given curve is

y = $$2x^2 + 3 sin x$$                   ……….(i)

Putting x = 0 in (i), we get y = 0.

So, the point of contact is (0, 0).

Now, y = $$2x^2 + 3 sin x$$

Differentiating with respect to x,

$$\implies$$  $$dy\over dx$$  = 4x + 3 cos x

$$\implies$$  $$({dy\over dx})_{(0, 0)}$$ = 4 $$\times$$ 0 + 3 cos 0 = 3

So, the equation of the normal at (0, 0) is

y – 0 = -$$1\over 3$$(x – 0)  or,  x + 3y = 0

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