Find the equation of the normal to the curve 2x²= y, which passes through (1, 2)

Question

Question

Answer 1

Concept:

Slope of tangent to the curve = \(\rm \frac {dy}{dx}\)

Slope of normal to the curve = \(\rm \frac{-1}{(\frac {dy}{dx})}\)

Point-slope is the general form: y - y₁ = m(x - x₁),

Where m = slope

Calculation:

Here, y = 2x²

\(\rm \frac {dy}{dx}\) = 4x

\(\rm \frac{dy}{dx}|_{x=1}=4\)

Slope of normal to the curve =\(\rm \frac{-1}{(\frac {dy}{dx})}\) = -1/4

Equation of normal to curve passing through (1, 2) is

(y - 2) = (-1/4)(x - 1)

⇒ 4y - 8 = -x + 1

⇒ x + 4y - 9 = 0

Hence, option (4) is correct.

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