Parabola Question 2
Question 2 - 2024 (29 Jan Shift 2)
Let $\mathrm{P}(\alpha, \beta)$ be a point on the parabola $\mathrm{y}^{2}=4 \mathrm{x}$. If $\mathrm{P}$ 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
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Answer (192)
Solution
Parabola is $x^{2}=8 y$
Chord with mid point $\left(\mathrm{x}{1}, \mathrm{y}{1}\right)$ is $\mathrm{T}=\mathrm{S}_{1}$
$\therefore \mathrm{xx}{1}-4\left(\mathrm{y}+\mathrm{y}{1}\right)=\mathrm{x}{1}^{2}-8 \mathrm{y}{1}$
$\therefore\left(\mathrm{x}{1}, \mathrm{y}{1}\right)=\left(1, \frac{5}{4}\right)$
$\Rightarrow \mathrm{x}-4\left(\mathrm{y}+\frac{5}{4}\right)=1-8 \times \frac{5}{4}=-9$
$\therefore \mathrm{x}-4 \mathrm{y}+4=0$…
$(\alpha, \beta)$ lies on (i) $\backslash &$ also on $y^{2}=4 x$
$\therefore \alpha-4 \beta+4=0 \ldots(i i)$
$& \beta^{2}=4 \alpha \ldots$. (iii)
Solving (ii) $\backslash$ & (iii)
$\beta^{2}=4(4 \beta-4) \Rightarrow \beta^{2}-16 \beta+16=0$
$\therefore \beta=8 \pm 4 \sqrt{3}$ and $\alpha=4 \beta-4=28 \pm 16 \sqrt{3}$
$\therefore(\alpha, \beta)=(28+16 \sqrt{3}, 8+4 \sqrt{3}) &$
$(28-16 \sqrt{3}, 8-4 \sqrt{3})$
$\therefore(\alpha-28)(\beta-8)=( \pm 16 \sqrt{3})( \pm 4 \sqrt{3})$
$=192$
