QJEEM24Ma4159: JEE Main 2024 Mathematics - Class 11 JEEM24Ma4159 Solution: Let be a point on the parabola . If also lies on the chord of…

Parabola is $x^2=8 y$

Chord with mid point $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is $\mathrm{T}=\mathrm{S}_1$

$\begin{aligned}& \therefore \mathrm{xx}_1-4\left(\mathrm{y}+\mathrm{y}_1\right)=\mathrm{x}_1^2-8 \mathrm{y}_1 \\& \therefore\left(\mathrm{x}_1, \mathrm{y}_1\right)=\left(1, \frac{5}{4}\right) \\& \Rightarrow \mathrm{x}-4\left(\mathrm{y}+\frac{5}{4}\right)=1-8 \times \frac{5}{4}=-9\end{aligned}$

$\therefore x-4 y+4=0$ ...... (i)

$(\alpha, \beta)$ lies on (i) & also on $y^2=4 x$

$\begin{aligned}& \therefore \alpha-4 \beta+4=0 \text{ .... (ii)} \\& \& ~\beta^2=4 \alpha \text{ .... (iii)}\end{aligned}$

Solving (ii) & (iii)

$\begin{aligned}& \beta^2=4(4 \beta-4) \Rightarrow \beta^2-16 \beta+16=0 \\& \therefore \beta=8 \pm 4 \sqrt{3} \text { and } \alpha=4 \beta-4=28 \pm 16 \sqrt{3} \\& \therefore(\alpha, \quad \beta) \quad=\quad(28+16 \sqrt{3}, 8+4 \sqrt{3}) \quad \& \\& (28-16 \sqrt{3}, 8-4 \sqrt{3}) \\& \therefore(\alpha-28)(\beta-8)=( \pm 16 \sqrt{3})( \pm 4 \sqrt{3}) \\& =192\end{aligned}$