Find the equation of the normal to the curve | vedclass

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(A) Given the curve $x^{2}=4y$. Differentiating with respect to $x$,we get $2x = 4 \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{x}{2}$. Let the

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$x+y-3=0$

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(A) Given the curve $x^{2}=4y$. Differentiating with respect to $x$,we get $2x = 4 \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{x}{2}$. Let the point of contact be $(h, k)$. Since the point lies on the curve,$h^{2} = 4k$,so $k = \frac{h^{2}}{4}$. The slope of the tangent at $(h, k)$ is $m_{T} = \frac{h}{2}$. The slope of the normal at $(h, k)$ is $m_{N} = -\frac{2}{h}$. The equation of the normal at $(h, k)$ is $y - k = -\frac{2}{h}(x - h)$. Since the normal passes through $(1, 2)$,we substitute $x=1$ and $y=2$: $2 - \frac{h^{2}}{4} = -\frac{2}{h}(1 - h)$ $2 - \frac{h^{2}}{4} = -\frac{2}{h} + 2$ $-\frac{h^{2}}{4} = -\frac{2}{h}$ $h^{3} = 8 \Rightarrow h = 2$. If $h = 2$,then $k = \frac{2^{2}}{4} = 1$. The slope of the normal is $m_{N} = -\frac{2}{2} = -1$. The equation of the normal is $y - 1 = -1(x - 2)$,which simplifies to $y - 1 = -x + 2$,or $x + y - 3 = 0$.

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

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