The normal to the curve x^{2}=4y passing thro | vedclass

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(D) The equation of the given curve is $x^{2}=4y$.Differentiating with respect to $x$,we have:$2x = 4 \cdot \frac{dy}{dx}$$\Rightarrow \frac{dy}{dx} =

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$x+y=3$

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(D) The equation of the given curve is $x^{2}=4y$.Differentiating with respect to $x$,we have:$2x = 4 \cdot \frac{dy}{dx}$$\Rightarrow \frac{dy}{dx} = \frac{x}{2}$.The slope of the normal to the curve at point $(h, k)$ is given by:$m = -\frac{1}{\left(\frac{dy}{dx}\right)_{(h,k)}} = -\frac{2}{h}$.The equation of the normal at point $(h, k)$ is:$y - k = -\frac{2}{h}(x - h)$.Since the normal passes through $(1, 2)$,we have:$2 - k = -\frac{2}{h}(1 - h) \Rightarrow k = 2 + \frac{2}{h}(1 - h) \dots (i)$.Since $(h, k)$ lies on the curve $x^{2} = 4y$,we have $h^{2} = 4k$,which implies $k = \frac{h^{2}}{4}$.Substituting $k$ in equation $(i)$:$\frac{h^{2}}{4} = 2 + \frac{2}{h} - 2$$\frac{h^{2}}{4} = \frac{2}{h}$$h^{3} = 8 \Rightarrow h = 2$.Then $k = \frac{2^{2}}{4} = 1$.The equation of the normal at $(2, 1)$ is:$y - 1 = -\frac{2}{2}(x - 2)$$y - 1 = -(x - 2)$$y - 1 = -x + 2$$x + y = 3$.The correct answer is $D$.

The normal to the curve $x^{2}=4y$ passing through $(1,2)$ is

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