Q.Two tangents to the parabola $y^2 = 8x$ meet the tangents at the vertex in the point P and Q. If PQ = 4, prove that the equation of the locus of the point of intersection of two tangent is $y^2 = 8(x + 2)$.

A tangent to a parabola is the limiting position of a secant line as its two points of intersection with the curve merge into one — equivalently, a line touching the curve at exactly one point without crossing it there. Differentiating the standard equation $y^2=4ax$ gives the slope $dy/dx=2a/y$ at any point, which leads to the point-form tangent at $(x_1,y_1)$: $yy_1=2a(x+x_1)$ — obtained from the original equation by the simple substitution $y^2\to yy_1$ and $x\to\frac{x+x_1}{2}$.

For a tangent of a given slope $m$ (rather than at a specific point), comparing the general line $y=mx+c$ against the point-form tangent yields the condition of tangency: the line touches the parabola exactly when $c=a/m$, and the tangent can then be written directly as $y=mx+\dfrac{a}{m}$, touching at the point $\left(\dfrac{a}{m^2},\dfrac{2a}{m}\right)$. …