Mathematics · Ellipse

JEE Main 2025 — 24 January, Evening Shift — Question 1

The equation of the chord, of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$ , whose mid-point is $(3,1)$ is :

Answer: A

Equation of chord with given middle point

$\mathrm{T}=\mathrm{S}$ ,

$\Rightarrow \frac{3 \mathrm{x}}{25}+\frac{\mathrm{y}}{16}-1=\frac{9}{25}+\frac{1}{16}-1$

$48 \mathrm{x}+25 \mathrm{y}=144+25$

$48 \mathrm{x}+25 \mathrm{y}=169$

Answer key and solution verified before publishing.

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