If the tangents drawn at the points P and Q on the parabola y ^ 2 =...

On plotting the diagram of given value we get,

Now given parabola is ${y}^{2} = 2x - 3 ⇒ {y}^{2} = 2\left(x - \dfrac{3}{2}\right)$

Let a point on parabola be $P = \left(\left(\dfrac{3}{2} + \dfrac{{t}^{2}}{2}\right), t\right)$

Equation of tangent at $P$ will be $yt = x + \dfrac{3}{2} + \dfrac{{t}^{2}}{2} - 3$

$∵ R\left(0, 1\right)$ lies on this tangent

Then $t = \dfrac{{t}^{2}}{2} - \dfrac{3}{2}$

i.e. $t = 3$ or $- 1$.

So $P ≡ \left(6, 3\right) \text{ and } Q ≡ \left(2, - 1\right)$

Now triangle $PQR$ is right angled at $Q$, so \left(2, - 1\right) is the orthocentre of the triangle.