I am working through a pure maths book as a hobby. I am tackling this problem.
$P(p^2,p^3)$ and $Q(q^2,q^3)$ are two points on the curve $y^2=x^3$. Find the equation of the chord PQ and deduce the equation of the tangent at P. Given that the tangent at P passes through Q and is normal to the curve at Q, find the values of p and q.
I have said gradient of chord =:
$\frac{q^3-p^3}{q^2-p^2} = \frac{q^2+pq+p^2}{p+q}$
$\rightarrow \frac{y-p^3}{x-p^2} = \frac{q^2+pq+p^2}{p+q}$
$\rightarrow y(p+q) = x(x^2+pq+p^2)-p^2q^2$
For equation of tangent let $q \rightarrow p$
Then $2py=x3p^2-p^4 \rightarrow 2y=3px-p^3$
But the last part has me stumped. I imagine the scenario as illustrated below but can't think of a way to find p and q.
Given curve is $y^2 = x^3$. Slope of its tangent is given by,
$\dfrac{dy}{dx} = \dfrac{3x^2}{2y}$
Now as you found, the gradient of line $PQ$ is $\dfrac{p^2+pq+q^2}{p+q}$.
But we also know that $PQ$ is tangent to the curve at point $P$ and is normal to the curve at point $Q$.
Slope of tangent at point $P$ is $\dfrac{3 p^4}{2p^3} = \dfrac{3p}{2}$ (for $p \ne 0)$
Slope of normal line to the curve at $Q$ is then $-\dfrac{2}{3q}$ (for $q \ne 0$)
Equating $\dfrac{p^2+pq+q^2}{p+q} = \dfrac{3p}{2} = -\dfrac{2}{3q} \ $ and solving,
$p = \pm \dfrac{2\sqrt2}{3}, q = \mp \dfrac{\sqrt2}{3}$
This gives you two such pairs of points $P, Q$ as shown in your diagram.
Edit: to your later question on steps to solve the equations -
$\dfrac{p^2+pq+q^2}{p+q} = \dfrac{3p}{2} = -\dfrac{2}{3q}$
From $\dfrac{3p}{2} = -\dfrac{2}{3q}, \ pq = - \dfrac{4}{9} \ $...$(i)$
From $\dfrac{p^2+pq+q^2}{p+q} = \dfrac{3p}{2}, \ p^2 - 2q^2 + pq = 0 \ $ ...$(ii)$
Substituting $q$ from $(i)$ in $(ii)$,
$p^2 - 2 \left(-\dfrac{4}{9p}\right)^2 - \dfrac{4}{9} = 0$
Simplifying, $81p^4 - 36p^2 - 32 = 0$
Completing the square, $(9p^2 - 2)^2 = 36$
$9p^2 = 2 \pm 6$ and given $p$ is real, we get $p = \pm \dfrac{2\sqrt2}{3}$. Now you can easily find $q$ from equation $(i)$.