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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 :

  1. A  \(\frac{\sqrt{1691}}{5}\)
  2. B  \(\frac{\sqrt{2009}}{5}\)
  3. C  \(\frac{\sqrt{1741}}{5}\)
  4. D  \(\frac{\sqrt{1541}}{5}\)
Verified Solution

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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