If the normal to the parabola y^2=4x at P(1,2 | vedclass

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(B) The given equation of the parabola is $y^2=4x$. For a point $P(x_1, y_1)$ on the parabola $y^2=4ax$,the normal at $P$ is given by $y-y_1 = -\frac{

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$(9,-6)$

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(B) The given equation of the parabola is $y^2=4x$. For a point $P(x_1, y_1)$ on the parabola $y^2=4ax$,the normal at $P$ is given by $y-y_1 = -\frac{y_1}{2a}(x-x_1)$. Here,$a=1$ and $P=(1,2)$,so the slope of the normal is $m = -\frac{2}{2(1)} = -1$. The equation of the normal at $P(1,2)$ is $y-2 = -1(x-1)$,which simplifies to $x+y=3$,or $x=3-y$. Substituting $x=3-y$ into the parabola equation $y^2=4x$: $y^2 = 4(3-y)$ $y^2 = 12-4y$ $y^2+4y-12=0$ $(y-2)(y+6)=0$ This gives $y=2$ (which corresponds to point $P$) and $y=-6$. For $y=-6$,$x = 3-(-6) = 9$. Therefore,the coordinates of point $Q$ are $(9,-6)$.

If the normal to the parabola $y^2=4x$ at $P(1,2)$ meets the parabola again at $Q$,then the coordinates of $Q$ are

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