Find the equation of normal to the curve $x^{2} = 4 y$ passing through the point $(1, 2)$

x + y = 3

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x - y = 3

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2 x - y = 4

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2 x - 3 y = 1

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Solution

The correct option is A $x + y = 3$
$x^{2} = 4 y$

$\Rightarrow \frac{d y}{d x} = \frac{x}{2} = m$ (say)
Thus slope of normal at $(1, 2)$ is $m^{'} = - \frac{1}{m} = \frac{- 2}{x} = \frac{- 2}{2} = - 1$

Hence, equation of normal at $(1, 2)$ to the parabola is,

$(y - 2) = - 1 (x - 1)$

$\Rightarrow x + y = 3$

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Similar questions

The normal to the curve x² = 4y passing (1, 2) is

(A) x + y = 3 (B) x − y = 3

y = x^{3} - 2 x + 4

(1, 3)

List-I	List-II
a) Equation of tangent to the curve $y = b e^{- x / a}$ at $x = 0$	1) $x - 2 y = 2$
b) Equation of tangent to the curve $y = x^{2} + 1$ at $(1, 2)$	2) $y = 2 x$
c) Equation of normal to the curve $y = 2 x - x^{2}$ at $(2, 0)$	3) $x - y = π$
d) Equation of normal to the curve $y = sin x$ at $x = π$	4) $\frac{x}{a} + \frac{y}{b} = 1$

(x, y)

{tan}^{- 1} (2 x + 3 y)

(1, 2)

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