Mathematics · Parabola

JEE Main 2024 — 9 April, Shift 2 — Question 21

Consider the circle C:x2+y2=4C: x^{2}+y^{2}=4 and the parabola P:y2=8xP: y^{2}=8 x.If the set of all values of α\alpha, for which three chords of the circle C on three distinct lines passing through the point (α,0)(\alpha, 0) are bisected by the parabola PP is the interval (p,q)(p, q), then (2q−p)2(2 q-p)^{2} is equal to \qquad

Answer: 80

Numerical answer — enter this value.

Step-by-step solution

figure

T=S1\mathrm{T}=\mathrm{S}_{1}

xx1+yy1=x12+y12\mathrm{xx}_{1}+\mathrm{yy}_{1}=\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}

αx1=x12+y12\alpha \mathrm{x}_{1}=\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}

α(2t2)=4t4+16t2\alpha\left(2 \mathrm{t}^{2}\right)=4 \mathrm{t}^{4}+16 \mathrm{t}^{2}

α=2t2+8\alpha=2 \mathrm{t}^{2}+8

α−82=t2\frac{\alpha-8}{2}=\mathrm{t}^{2}

Also, 4t4+16t2−4<04 \mathrm{t}^{4}+16 \mathrm{t}^{2}-4<0

t2=−2+5t^{2}=-2+\sqrt{5}

α=4+25\alpha=4+2 \sqrt{5}

∴α∈(8,4+25)\therefore \alpha \in(8,4+2 \sqrt{5})

∴(2q−p)2=80\therefore(2 \mathrm{q}-\mathrm{p})^{2}=80

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Parabola
Topic
Introduction to Parabola
Consider the circle C: x 2 +y 2 =4 and the parabola P: y 2 =8 x .If… | JEE Main 2024 PYQ with Solution · DhiX AI