JEE MAIN - Mathematics (2025 - 24th January Evening Shift - No. 14)
The equation of the chord, of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, whose mid-point is $(3,1)$ is :
$5 x+16 y=31$
$48 x+25 y=169$
$4 x+122 y=134$
$25 x+101 y=176$
Explanation
$$\begin{aligned}
&\text { Equation of chord with given middle point }\\
&\begin{aligned}
& \mathrm{T}=\mathrm{S}_1 \\
& \Rightarrow \frac{3 \mathrm{x}}{25}+\frac{\mathrm{y}}{16}-1=\frac{9}{25}+\frac{1}{16}-1 \\
& 48 \mathrm{x}+25 \mathrm{y}=144+25 \\
& 48 \mathrm{x}+25 \mathrm{y}=169 \text { Ans. }
\end{aligned}
\end{aligned}$$
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