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 $y^2=4x$.The equation of the tangent at $P(1,2)$ is $y(2) = 2(x+1)$,which simplifies to $y = x+1$.The slope

Vedclass · Accepted Answer

$(9,-6)$

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(B) The given equation of the parabola is $y^2=4x$.The equation of the tangent at $P(1,2)$ is $y(2) = 2(x+1)$,which simplifies to $y = x+1$.The slope of the tangent is $m = 1$.Therefore,the slope of the normal is $m' = -1/m = -1$.The equation of the normal passing through $P(1,2)$ is $y - 2 = -1(x - 1)$,which simplifies to $x + y = 3$,or $x = 3 - y$.Substituting this 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 $y = 2$ (at point $P$) and $y = -6$ (at point $Q$).For $y = -6$,$x = 3 - (-6) = 9$.Thus,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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