Question

2 tangents to a parabola y^2=8x meet the tangent at the vertex in the point p and q. If pq=4 prove that the locus of point of intersection of these two tangents is y^2=8(x+2)

Answer 1

Proved below.

Step-by-step explanation:

Given:

Here PQ = 4

Let the two tangents be $t_1y = x + 2t^2_1$ and $t_2y = x + 2t^2_2$

So, that their point of intersection is $[2t_1t_2, 2(t_1 + t_2)]$

Now,

$P = (0, 2t_1)$ and $Q = (0, 2t_2).$

Since PQ = 4, we have $|t_1 - t_2| = 2$          [1]

Suppose $x_1 = 2t_1t_2$ and $y_1 = 2(t_1 + t_2)$.    [2]

Therefore  

$\frac{y^2_1}{4}=(t_1+t_2)^2$

$\frac{y^2_1}{4}=t^2_1+2t_1t_2+t^2_2$

$\frac{y^2_1}{4}=t^2_1-2t_1t_2+t^2_2+4t_1t_2$

$\frac{y^2_1}{4}=(t_1-t_2)^2+4t_1t_2=4+2x_1$           [from Eq (1),(2)]

$\frac{y^2_1}{4}=2(2+x_1)$

Thus, the locus of $(x_1, y_1)$ is $y^2 = 8(x + 2)$.

Hence proved.