The tangents drawn to a parabola at points (1,2) and (3,4) intersect at (-2,-3). Prove that the slope of the axis of the parabola is $\frac{3}{2}$

Question

The vertex of the parabola is not given in the question. I have tried to solve this question by writing the general eqn of a tangent for the parabola in point form, and using a third equation $T^2=SS_1$ but it's very long and I couldn't solve it. Can someone suggest a shorter method.

Answer 1

Let $P \equiv (1,2)$, $Q\equiv (3,4)$ then $PQ$ is a chord of the parabola. Now, out of all the diameters(lines parallel to the axis) of this parabola, one will bisect this chord and all other chords parallel to $PQ$.

The diameter passing through the midpoint $R\equiv (\frac{1+3}{2},\frac{2+4}{2}) \equiv (2,3)$ of $PQ$ will also pass through the point of intersection $T \equiv (-2,-3)$ of tangents through $P$ and $Q$ (You can try to prove this statement). Hence, the slope of axis of parabola being parallel to this diameter $RT$ is $\frac{3-(3)}{2-(-2)}=\frac{3}{2}$.

Answer 2

Let $A$ and $B$ be the given points of tangency, and let $P$ be the intersection of the corresponding tangents. The diameter through $P$ must bisect chord $AB$, and every diameter is parallel to the axis.