Q.

Let y2=12x be the parabola and S be its focus. Let PQ be a focal chord of the parabola such that (SP)(SQ) = 1474. Let C be the circle described taking PQ as a diameter. If the equation of a circle C is 64x2+64y2–αx–643y=β, then β–α is equal to __________.          [2025]


Ans.

(1328)

Given, equation of parabola is y2=12x

Focus = S = (3, 0)

Let P(3t12,6t1) and Q(3t22,6t2) are points on parabola

Also, t1t2=–1

∴  P≡(3t2,6t) and Q≡(3t2,–6t)

Now, (SP)(SQ)=1474 ⇒ (3+3t2)(3+3t2)=1474

(∵ (SP) = PM and (SQ) = QN, where PM and QN are perpendicular distance from directrix)

⇒ (1+t2)2t2=4912 ⇒ 12t4–25t2+12=0

⇒ t2=34,43 ⇒ t=±32 or t=±23

Case I : When t=32 or t=–23

Points are P(94,33) and Q(4,–43)

∴   Equation of circle is

          (x–4)(x–94)+(y–33)(y+43)=0

⇒ x2+y2–25x4+3y–27=0

On comparing with given equation of circle 64x2+64y2–αx

–643y=β, we get α = 400, β = 1728

Case II : When t=–32 or t=23

Point are P(94,–33) and Q(4,43)

Similarly, we get α = 400, β = 1728

∴  β–α = 1728 – 400 = 1328.