Q.

Let P(α, β) be a point on the parabola y2=4x. If P also lies on the chord of the parabola x2=8y whose mid point is (1, 54), then (α–28)(β–8) is equal to __________.          [2024]


Ans.

(192)

We α = t2, β = 2t

Now, equation of chord of the parabola x2=8y, bisected at (1, 54) is x·1–2×2(y+54)=1–10

⇒ x–4y–5=–9 ⇒ x–4y+4=0

Now, (α, β) satisfies the above equation

∴  t2–8t+4=0            ... (i)

Now, (α–28)(β–8)=(t2–28)(2t–8)

=(8t–4–28)(2t–8)           (Using (i))

=16t2–64t–64t+256=16(t2–8t+16)

=16(t2–8t+4+12)=192.