JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
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 \(5 x+16 y=31\)
- C \(25 x+101 y=176\)
- D \(4 x+122 y=134\)
Answer & Solution
Correct Answer
(A) \(48 x+25 y=169\)
Step-by-step Solution
Detailed explanation
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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