Parabola
Chord with Given Midpoint — Intersection with Another Parabola
nta_pyq_2024_jan
Grade 11
Question:
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 $\left(1,\dfrac{5}{4}\right)$, then $(\alpha-28)(\beta-8)$ is equal to
Step-by-Step Solution
Key Concept: Chord of $x^2=8y$ with midpoint $(x_1,y_1)=(1,5/4)$: use $T=S_1$: $xx_1-4(y+y_1)=x_1^2-8y_1$. Simplify to get the chord equation. Since $P(\alpha,\beta)$ lies on this chord and on $y^2=4x$, solve simultaneously.
Chord: $x-4y+4=0$. $\alpha=4\beta-4$, $\beta^2=4\alpha$: $\beta=8\pm4\sqrt3$, $\alpha=28\pm16\sqrt3$. $(\alpha-28)(\beta-8)=192$.
Correct Answer: 192