Let y^2=12 x be the parabola and S be its focus. Let PQ be a focal chord of the parabola such that (SP)(SQ)=(147)/(4) . Let C be the circle described taking PQ as a diameter. If the equation of a circle C is 64 x^2+64

Correct answer: 0 y^2=12 x a=3 SP × SQ=(147)/(4) Let P (3 t^2, 6 t) and t_1 t_2=-1 (ends of focal chord) So, Q((3)/(t^2), (-6)/(t)) aligned & S(3,0) & SP × SQ=PM_1 × QM_2 aligned (dist. from directrix) aligned & =(3+3 t^2)(3+ 3 t^2 )=(147)/(4) & (1+t^2)^2 t^2 =(49)/(12) & t^2=(3)/(4), (4)/(3) & t= ± 3 2 , ± 2 3 aligned considering t= - 3 2 P((9)/(4),-3 3 ) and Q(4,4 3 ) Hence, diametric circle: aligned & (x-4)(x-(9)/(4))+(y+3 3 )(y-4 3 )=0 & x^2+y^2-(25)/(4) x- 3 y-27=0 & α=400, β=1728 & β-α=1328 aligned

JEE Main 202529 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Parabola

JEE Main 2025 · 29 January · Evening Shift · Mathematics · Parabola

Let y^2=12 x be the parabola and S be its focus. Let PQ be a focal...

Let

y^2=12 x

be the parabola and

S

be its focus. Let PQ be a focal chord of the parabola such that

(\mathrm{SP})(\mathrm{SQ})=\frac{147}{4}

. Let C be the circle described taking PQ as a diameter. If the equation of a circle

C

64 x^2+64 y^2-\alpha x-64 \sqrt{3} y=\beta

, then

\beta-\alpha

is equal to ________.

Correct answer

Step-by-step solution

\mathrm{y}^2=12 \mathrm{x} \quad \mathrm{a}=3 \quad \mathrm{SP} \times \mathrm{SQ}=\frac{147}{4}

Let

\mathrm{P}

\left(3 \mathrm{t}^2, 6 \mathrm{t}\right)

and

\mathrm{t}_1 \mathrm{t}_2=-1

(ends of focal chord) So,

Q\left(\frac{3}{t^2}, \frac{-6}{t}\right)

\begin{aligned} & \mathrm{S}(3,0) \\ & \mathrm{SP} \times \mathrm{SQ}=\mathrm{PM}_1 \times \mathrm{QM}_2 \end{aligned}

(dist. from directrix)

\begin{aligned} & =\left(3+3 \mathrm{t}^2\right)\left(3+\frac{3}{\mathrm{t}^2}\right)=\frac{147}{4} \\ & \Rightarrow \frac{\left(1+\mathrm{t}^2\right)^2}{\mathrm{t}^2}=\frac{49}{12} \\ & \mathrm{t}^2=\frac{3}{4}, \frac{4}{3} \\ & \mathrm{t}= \pm \frac{\sqrt{3}}{2}, \pm \frac{2}{\sqrt{3}} \end{aligned}

considering

\mathrm{t}=\frac{-\sqrt{3}}{2}

\mathrm{P}\left(\frac{9}{4},-3 \sqrt{3}\right) \text{ and } \mathrm{Q}(4,4 \sqrt{3})

Hence, diametric circle:

\begin{aligned} & (x-4)\left(x-\frac{9}{4}\right)+(y+3 \sqrt{3})(y-4 \sqrt{3})=0 \\ & \Rightarrow x^2+y^2-\frac{25}{4} x-\sqrt{3} y-27=0 \\ & \Rightarrow \alpha=400, \beta=1728 \\ & \beta-\alpha=1328 \end{aligned}

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