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10-2. Parabola, Ellipse, Hyperbola
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The equation of the chord, of the ellipse $\frac{ x ^2}{25}+\frac{ y ^2}{16}=1$, whose mid-point is $(3,1)$ is :
A$48 x+25 y=169$
B$4 x+122 y=134$
C$25 x+101 y=176$
D$5 x+16 y=31$
(JEE MAIN-2025)
Solution
Equation of chord with given middle point
$T=S_1$
$\Rightarrow \frac{3 x}{25}+\frac{y}{16}-1=\frac{9}{25}+\frac{1}{16}-1$
$48 x+25 y=144+25$
$48 x+25 y=169 \text { Ans. }$
$T=S_1$
$\Rightarrow \frac{3 x}{25}+\frac{y}{16}-1=\frac{9}{25}+\frac{1}{16}-1$
$48 x+25 y=144+25$
$48 x+25 y=169 \text { Ans. }$
Standard 11
Mathematics