The line y = x + 1 meets the ellipse x ^ 2 4 + y ^ 2 2 = 1 at two...

Plotting the diagram of given data we have,

Let $P$ be the point \left(- 1 + \dfrac{p}{\sqrt{2}}, \dfrac{p}{\sqrt{2}}\right)$</p><p>Since it lies on the ellipse$\dfrac{{x}^{2}}{4} + \dfrac{{y}^{2}}{2} = 1$, we get </p><p>$\dfrac{{\left(- 1 + \dfrac{p}{\sqrt{2}}\right)}^{2}}{4} + \dfrac{{\left(\dfrac{p}{\sqrt{2}}\right)}^{2}}{2} = 1 ⇒ \dfrac{3{p}^{2}}{8} - \dfrac{p}{2\sqrt{2}} - \dfrac{3}{4} = 0$</p><p>Here${\left({p}_{1} - {p}_{2}\right)}^{2} = {\left(\dfrac{2\sqrt{2}}{3}\right)}^{2} - 4\left(- 2\right) = \dfrac{8}{9} + 8 = \dfrac{80}{9}$</p><p>We know,${\left({p}_{1} - {p}_{2}\right)}^{2} = P{Q}^{2}$</p><p>$⇒ P{Q}^{2} = \dfrac{80}{9}$</p><p>Now$r = \dfrac{PQ}{2} ⇒ 9{r}^{2} = 9\dfrac{{\left(PQ\right)}^{2}}{4} = \dfrac{9}{4} \times \dfrac{80}{9} = 20