Two tangents to the parabola $y^2= 8x$ meet the tangent at its vertex in the points $P$ and $Q$. If $|PQ| = 4$, prove that the locus of the point of the intersection of the two tangents is $y^2 = 8 (x + 2)$. I take the point of intersection $(x,y)$ . But after that how can I write equation of the tangents?
Use the fact that perpendicular from the focus on any tangent to a parabola meets the tangent at the vertex. So let$ty=x+at^2$ be the equation of tangent. Equation of line perpendicular from focus is $y=-tx+ta$ Solve it with the equation of tangent. Get the points. Use the fact that the distance is given and the point of intersection of tangents is $(at_1t_2, a(t_1+t_2)$
This is to separate the geometric statement implied in the problem from the particular numbers used.
The question claims that if we slide a line segment of constant length $L$ along a parabola, the locus of the intersection point of the tangents at the ends of the segment is another parabola. In addition, the second parabola is a translation of the first along the axis of symmetry. The magnitude of the translation is apparently some simple function of $L$ that sends $2$ to $4$.