JEE Main 2024 (Online) 9th April Evening ShiftQuestion 1
Consider the circle $C:x^{2}+y^{2}=4$ and the parabola $P:y^{2}=8x$. If the set of all values of $\alpha$, for which three chords of the circle $C$ on three distinct lines passing through the point $(\alpha,0)$ are bisected by the parabola $P$ is the interval $(p,q)$, then $(2q-p)^{2}$ is equal to $\_\_\_\_\_\_$ .
Parabola
Answer$80$
Solution
Concept. Take the midpoint $(h,k)$ of a chord of the circle on the parabola; the chord must pass through $(\alpha,0)$ and the midpoint must be inside the circle. Formulas used. (1) Chord of $x^2+y^2=4$ with midpoint $(h,k)$: $hx+ky=h^2+k^2$.(2) Midpoint on $y^2=8x$: $k^2=8h$.(3) Midpoint strictly inside the circle: $h^2+k^2<4$.Step 1 - Through $(\alpha,0)$. From (1): $h\alpha=h^2+k^2=h^2+8h$. So $h=0$ or $h=\alpha-8$. Step 2 - Case h = 0. Then $k=0$: midpoint is the centre, the chord through $(\alpha,0)$ is the $x$-axis. One line. Step 3 - Case h = α - 8 > 0. $k=\pm\sqrt{8h}$ gives two more lines. Use (3): $$h^2+8h<4\Rightarrow(h+4)^2<20\Rightarrow h<2\sqrt5-4.$$ Step 4 - Range. $0<\alpha-8<2\sqrt5-4\Rightarrow\alpha\in(8,\,4+2\sqrt5)$. $p=8$, $q=4+2\sqrt5$. Step 5. $2q-p=4\sqrt5$, $(2q-p)^2=80$. Answer: 80