QJEEM22Ma4c87: JEE Main 2022 Mathematics - Class 11 JEEM22Ma4c87 Solution: The line y = x + 1 meets the ellipse at two points P and Q. If…

Let point (a, a + 1) as the point of intersection of line and ellipse.

So, ${{{a^2}} \over 4} + {{{{(a + 1)}^2}} \over 2} = 1 \Rightarrow {a^2} + 2({a^2} + 2a + 1) = 4$

$\Rightarrow 3{a^2} + 4a - 2 = 0$

If roots of this equation are $\alpha$ and $\beta$ .

So, $P(\alpha ,\,\alpha + 1)$ and $Q(\beta ,\,\beta + 1)$

$PQ = 4{r^2} = {(\alpha - \beta )^2} + {(\alpha - \beta )^2}$

$\Rightarrow 9{r^2} = {9 \over 4}(2{(\alpha - \beta )^2})$

$= {9 \over 2}\left[ {{{(\alpha + \beta )}^2} - 4\alpha \beta } \right]$

$= {9 \over 2}\left[ {{{\left( { - {4 \over 3}} \right)}^2} + {8 \over 3}} \right]$

$= {1 \over 2}[16 + 24] = 20$