Ellipse | JEE Main 2024 Solution

JEE MAIN 2024

27-01-2024 SHIFT-1

Question

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 :

Select the correct option:

$\frac{\sqrt{1691}}{5}$

$\frac{\sqrt{2009}}{5}$

$\frac{\sqrt{1741}}{5}$

$\frac{\sqrt{1541}}{5}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Mathematics

Medium

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