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Find the equation of the normal to the curve 2x2 = y, which passes through (1, 2)
1. x + y + 9 = 0
2. 4x + y - 9 =0
3. 3x + 4y - 8 = 0
4. x + 4y - 9 = 0 
5. None of these

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Correct Answer - Option 4 : x + 4y - 9 = 0 

Concept:

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

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

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

Calculation:

Here, y = 2x2

\(\rm \frac {dy}{dx}\) = 4x

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

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