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

Question

Let $$P(\alpha, \beta)$$ be a point on the parabola $y^2=4 x$. If $P$ also lies on the chord of the parabola $x^2=8 y$ whose mid point is $$\left(1, \frac{5}{4}\right)$$, then $$(\alpha-28)(\beta-8)$$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 192

Accepted Answer

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)

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

Solutions · 1