Let P(α, β) be a point on the parabola y^2 = 4x. If P also lies on the chord of the parabola

Question

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 \(\mathrm{x}^{2}=8 \mathrm{y}\) whose mid point is \(\left(1, \frac{5}{4}\right)\), then \((\alpha-28)(\beta-8)\) is equal to _____.

Answer 1

Correct answer: 192

Parabola is \(x^{2}=8 y\)

Chord with mid point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is \(\mathrm{T}=\mathrm{S}_{1}\)

\(\therefore \mathrm{xx}_{1}-4\left(\mathrm{y}^{+} \mathrm{y}_{1}\right)=\mathrm{x}_{1}^{2}-8 \mathrm{y}_{1}\)

\(\therefore\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=\left(1, \frac{5}{4}\right)\)

\(\Rightarrow x-4\left(y+\frac{5}{4}\right)=1-8 \times \frac{5}{4}=-9\)

\(\therefore \mathrm{x}-4 \mathrm{y}+4=0 \quad ....(i)\)

\((\alpha, \beta)\) lies on (i) & also on \(y^{2}=4 x\)

\(\therefore \alpha-4 \beta+4=0 \quad....(ii)\)

& \( \beta^{2}=4 \alpha \quad ....(iii)\)

Solving (ii) & (iii)

\( \beta^{2}=4(4 \beta-4) \Rightarrow \beta^{2}-16 \beta+16=0\)

\(\therefore \beta=8 \pm 4 \sqrt{3} \text { and } \alpha=4 \beta-4=28 \pm 16 \sqrt{3} \)

\(\therefore(\alpha, \beta)=(28+16 \sqrt{3}, 8+4 \sqrt{3}) \ \&\ (28-16 \sqrt{3}, 8-4 \sqrt{3})\)

\( \therefore|(\alpha-28)(\beta-8)|=|16 \sqrt{3} \times 4 \sqrt{3}| \)

\( =192\)