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JEE Main 202524 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Ellipse

JEE Main 2025 · 24 January · Evening Shift · Mathematics · Ellipse

The equation of the chord, of the ellipse (x^2)/(25)+(y^2)/(16)=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 :

Options

A $48 x+25 y=169$
B $5 x+16 y=31$
C $25 x+101 y=176$
D $4 x+122 y=134$

Correct answer

A. $48 x+25 y=169$

Step-by-step solution

Equation of chord with given middle point

\begin{aligned} & T=S_1 \\ & \Rightarrow \frac{3 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}

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