The line y = x + 1 meets the ellipse ${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$ at two points P and Q. If r is the radius of the circle with PQ as diameter then (3r)² is equal to :

<p>Let point (a, a + 1) as the point of intersection of line and ellipse.</p> <p>So, $${{{a^2}} \over 4} + {{{{(a + 1)}^2}} \over 2} = 1 \Rightarrow {a^2} + 2({a^2} + 2a + 1) = 4$$</p> <p>$\Rightarrow 3{a^2} + 4a - 2 = 0$</p> <p>If roots of this equation are $\alpha$ and $\beta$.</p> <p>So, $P(\alpha ,\,\alpha + 1)$ and $Q(\beta ,\,\beta + 1)$</p> <p>$PQ = 4{r^2} = {(\alpha - \beta )^2} + {(\alpha - \beta )^2}$</p> <p>$\Rightarrow 9{r^2} = {9 \over 4}(2{(\alpha - \beta )^2})$</p> <p>$= {9 \over 2}\left[ {{{(\alpha + \beta )}^2} - 4\alpha \beta } \right]$</p> <p>$$ = {9 \over 2}\left[ {{{\left( { - {4 \over 3}} \right)}^2} + {8 \over 3}} \right]$$</p> <p>$= {1 \over 2}[16 + 24] = 20$</p>