Let P(,) be a point on the parabola y^2=4x.

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 midpoint is $\left(1,\dfrac54\right)$, then $(\alpha-28)(\beta-8)$ is equal to \_\_\_\_\_\_\_\_.

Answer: 192

Accepted Answer

Chord of $x^2=8y$ with midpoint $(1,5/4)$: taking endpoints $(x_A,x_A^2/8),(x_B,x_B^2/8)$ with $x_A+x_B=2$ and $x_A^2+x_B^2=20$ gives $x_Ax_B=-8$ and slope $=\dfrac{x_A+x_B}{8}=\dfrac14$.

$$y-\frac54=\frac14(x-1) \;\Rightarrow\; x-4y+4=0.$$

$P(\alpha,\beta)$ on both $y^2=4x$ and this line: $\alpha=4\beta-4$, so $\beta^2=4\alpha=16\beta-16\Rightarrow \beta^2-16\beta+16=0$.

$$(\alpha-28)(\beta-8)=4(\beta-8)^2=4(\beta^2-16\beta+64)=4(-16+64)=192.$$

Solutions · 1