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Mathematics · 2025

JEE Main · 29 January 2025, Shift 2 · Q23

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…

Let $\displaystyle y^2=12 x$ be the parabola and S be its focus. Let PQ be a focal chord of the parabola such that $\displaystyle (\mathrm{SP})(\mathrm{SQ})=\frac{147}{4}$. Let C be the circle described taking PQ as a diameter. If the equation of a circle C is $\displaystyle 64 x^2+64 y^2-\alpha x-64 \sqrt{3} y=\beta$, then $\displaystyle \beta-\alpha$ is equal to $\displaystyle \_\_\_\_$.
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.