Verified Solution • 2025–26 Session moderate

QJEEM24Mae6ac: JEE Main 2024 Mathematics - Class 11 JEEM24Mae6ac Solution: Consider the circle and the parabola . If the set of all values …

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 CC 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 __________.

Step-by-Step Verified Solution

JEE Main 2024 (Online) 9th April Evening Shift Mathematics - Parabola Question 18 English Explanation

Chord with the middle point (α,0)(\alpha, 0)

amp;⇒T=S1amp;⇒yy1−4(x+x1)=y12−8x1amp;⇒−4(x+α)=0−8αamp;⇒x+α=2α⇒x=α\begin{aligned}& \Rightarrow T=S_1 \\& \Rightarrow y y_1-4\left(x+x_1\right)=y_1^2-8 x_1 \\& \Rightarrow-4(x+\alpha)=0-8 \alpha \\& \Rightarrow x+\alpha=2 \alpha \Rightarrow x=\alpha\end{aligned}

For circle chord with (2t2,4t)(2 t^2, 4 t) as mid point

amp;⇒T=S1amp;⇒xx1+yy1−4=x12+y12−4amp;⇒2t2x+4ty=4t4+16t2\begin{aligned}& \Rightarrow \quad T=S_1 \\& \Rightarrow \quad x x_1+y y_1-4=x_1^2+y_1^2-4 \\& \Rightarrow \quad 2 t^2 x+4 t y=4 t^4+16 t^2\end{aligned}

Passes through (α,0)(\alpha, 0)

⇒amp;2t2α=4t4+16t2⇒amp;2α=4t2+16⇒α=2t2+8=x0+8amp;x2+y2=4 and y2=8x⇒amp;x2+8x−4=0⇒x0=−8+802⇒amp;p=8 and q=4+802⇒(2q−p)2=80\begin{aligned}\Rightarrow & 2 t^2 \alpha=4 t^4+16 t^2 \\\Rightarrow & 2 \alpha=4 t^2+16 \Rightarrow \alpha=2 t^2+8=x_0+8 \\& x^2+y^2=4 \text { and } y^2=8 x \\\Rightarrow & x^2+8 x-4=0 \Rightarrow x_0=\frac{-8+\sqrt{80}}{2} \\\Rightarrow & p=8 \text { and } q=4+\frac{\sqrt{80}}{2} \Rightarrow(2 q-p)^2=80\end{aligned}

Related Learning Resources & Exam Preparation