The equation of a tangent to the parabola y^2 | vedclass

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(D) The given tangent is $y = x + 2$,which is of the form $y = mx + c$ where $m = 1$ and $c = 2$.For a parabola $y^2 = 4ax$,the condition for $y = mx

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$(-2, 0)$

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(D) The given tangent is $y = x + 2$,which is of the form $y = mx + c$ where $m = 1$ and $c = 2$.For a parabola $y^2 = 4ax$,the condition for $y = mx + c$ to be a tangent is $c = a/m$.Here,$4a = 8$,so $a = 2$. Checking the condition: $c = 2/1 = 2$,which is satisfied.The locus of the intersection of two perpendicular tangents to a parabola is its directrix.The directrix of the parabola $y^2 = 8x$ is $x = -a$,which is $x = -2$.We are looking for a point on the line $y = x + 2$ that also lies on the directrix $x = -2$.Substituting $x = -2$ into the line equation: $y = -2 + 2 = 0$.Thus,the required point is $(-2, 0)$.

The equation of a tangent to the parabola $y^2 = 8x$ is $y = x + 2$. If another tangent is drawn to the parabola from a point on this line such that it is perpendicular to the given tangent,find the point.

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