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

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Question

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

Vedclass · Accepted Answer

$x+y=3$

Vedclass · Answer

(C) The equation of the given curve is $x^{2}=4y$.Differentiating with respect to $x$,we get:$2x = 4 \cdot \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{x}{2}$.The slope of the tangent at any point $(h, k)$ on the curve is $\frac{h}{2}$.Therefore,the slope of the normal at $(h, k)$ is $-\frac{2}{h}$.The equation of the normal at $(h, k)$ is:$y - k = -\frac{2}{h}(x - h)$.Since the normal passes through $(1, 2)$,we have:$2 - k = -\frac{2}{h}(1 - h) \Rightarrow 2 - k = -\frac{2}{h} + 2 \Rightarrow k = \frac{2}{h}$.Since $(h, k)$ lies on the curve $x^{2} = 4y$,we have $h^{2} = 4k$.Substituting $k = \frac{2}{h}$ into the equation $h^{2} = 4k$:$h^{2} = 4 \left(\frac{2}{h}\right) \Rightarrow h^{3} = 8 \Rightarrow h = 2$.Now,find $k$:$k = \frac{2}{h} = \frac{2}{2} = 1$.The point of contact is $(2, 1)$.The slope of the normal is $m = -\frac{2}{h} = -\frac{2}{2} = -1$.The equation of the normal is:$y - 1 = -1(x - 2)$$y - 1 = -x + 2$$x + y = 3$.Thus,the correct option is $C$.

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

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