Consider the parabola with vertex ((1)/(2), (3)/(4)) and the...

Equation of the parabola is ${\left(x - \dfrac{1}{2}\right)}^{2} = \left(y - \dfrac{3}{4}\right) . . . \left(1\right)$

Put $x = \dfrac{- 1}{2}$ in equation \left(1\right)$, we get$y = \dfrac{7}{4}$</p><p>Hence, point$P$is$P\left(\dfrac{- 1}{2}, \dfrac{7}{4}\right)$</p><p>Now, differentiate equation \left(1\right)$ on both sides, we get ${y}^{'} = 2\left(x - \dfrac{1}{2}\right)$

At point $P\left(\dfrac{- 1}{2}, \dfrac{7}{4}\right)$, the slope of tangent is $- 2$.

Therefore, the slope of normal is $\dfrac{1}{2}$

Equation of normal at $P\left(\dfrac{- 1}{2}, \dfrac{7}{4}\right)$ is $x = 2y - 4 . . . \left(2\right)$

On solving equations \left(1\right) \text{ and } \left(2\right)$, we get the intersection point$Q\left(2, 3\right)$</p><p>$∴ ( PQ{)}^{2} = {\left(2 + \dfrac{1}{2}\right)}^{2} + {\left(3 - \dfrac{7}{4}\right)}^{2}$</p><p>$= \dfrac{25}{4} + \dfrac{25}{16} = \dfrac{125}{16}