JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The length of the chord of the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\), whose mid point is \(\left(1, \frac{2}{5}\right)\), is equal to :
- A \(\frac{\sqrt{1691}}{5}\)
- B \(\frac{\sqrt{2009}}{5}\)
- C \(\frac{\sqrt{1741}}{5}\)
- D \(\frac{\sqrt{1541}}{5}\)
Answer & Solution
Correct Answer
(A) \(\frac{\sqrt{1691}}{5}\)
Step-by-step Solution
Detailed explanation
Equation of chord with given middle point. \(\mathrm{T}=\mathrm{S}_1\) \( \frac{x}{25}+\frac{y}{40}=\frac{1}{25}+\frac{1}{100} \) \( \frac{8 x+5 y}{200}=\frac{8+2}{200}\) \(y=\frac{10-8 x}{5}\) \(.......(i)\) \(\frac{x^2}{25}+\frac{(10-8 x)^2}{400}=1\) (put in original equation)…
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