JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 28

Tangents to $y^2=4x$: $y=mx+\frac1m$. Through $(\alpha,\beta)$: $\alpha m^2-\beta m+1=0$.

Slopes $m$ and $4m$. $5m=\frac\beta\alpha$ and $4m^2=\frac1\alpha$, so $4\beta^2=25\alpha$.

On the ellipse. $25\alpha^2+4\beta^2=1$ gives $25\alpha^2+25\alpha=1$, so $100\alpha^2+100\alpha=4$.

Evaluate. $(10\alpha+5)^2=100\alpha^2+100\alpha+25=29$. And $16\beta^2+50=100\alpha+50=10(10\alpha+5)$, so $(16\beta^2+50)^2=100\cdot29=2900$.

Sum $29+2900=2929$.

Answer: 2929