JEE Main 2024 (Online) 29th January Evening ShiftQuestion 1
Let $P(\alpha,\beta)$ be a point on the parabola $y^{2}=4x$. If $P$ also lies on the chord of the parabola $x^{2}=8y$ whose mid point is $\left(1,\frac54\right)$, then $(\alpha-28)(\beta-8)$ is equal to ______.
Parabola
Answer$192$
Solution
Concept. Find the chord by $T=S_1$, then intersect it with $y^2=4x$. Formulas used. (1) For $x^2=4ay$: $T\equiv xx_1-2a(y+y_1)$, $S_1=x_1^2-4ay_1$.Step 1 - Use (1). $a=2$, $(x_1,y_1)=\big(1,\frac54\big)$: $$x-4\Big(y+\frac54\Big)=1-10\Rightarrow x-4y+4=0.$$ Step 2 - P on both. $\alpha=4\beta-4$ and $\beta^2=4\alpha=16\beta-16$, so $\beta^2-16\beta+16=0$. Step 3 - Solve. $(\beta-8)^2=64-16=48$, so $\beta-8=\pm4\sqrt3$. Step 4. $\alpha-28=4\beta-32=4(\beta-8)$. $$(\alpha-28)(\beta-8)=4(\beta-8)^2=4\cdot48=192.$$ Answer: 192