Ellipse Question 2
Question 2 - 2024 (27 Jan Shift 1)
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) $\frac{\sqrt{1691}}{5}$
(2) $\frac{\sqrt{2009}}{5}$
(3) $\frac{\sqrt{1741}}{5}$
(4) $\frac{\sqrt{1541}}{5}$
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Answer (1)
Solution
Equation of chord with given middle point.
$\mathrm{T}=\mathrm{S}_{1}$
$\frac{x}{25}+\frac{y}{40}=\frac{1}{25}+\frac{1}{100}$
$\frac{8 x+5 y}{200}=\frac{8+2}{200}$
$y=\frac{10-8 x}{5}$
$\frac{x^{2}}{25}+\frac{(10-8 x)^{2}}{400}=1 \quad$ (put in original equation)
$\frac{16 x^{2}+100+64 x^{2}-160 x}{400}=1$
$4 x^{2}-8 x-15=0$
$x=\frac{8 \pm \sqrt{304}}{8}$
$x_{1}=\frac{8+\sqrt{304}}{8} ; x_{2}=\frac{8-\sqrt{304}}{8}$
Similarly, $y=\frac{10-18 \pm \sqrt{304}}{5}=\frac{2 \pm \sqrt{304}}{5}$
$\mathrm{y}{1}=\frac{2-\sqrt{304}}{5} ; \mathrm{y}{2}=\frac{2+\sqrt{304}}{5}$
Distance $=\sqrt{\left(\mathrm{x}{1}-\mathrm{x}{2}\right)^{2}+\left(\mathrm{y}{1}-\mathrm{y}{2}\right)^{2}}$
$=\sqrt{\frac{4 \times 304}{64}+\frac{4 \times 304}{25}}=\frac{\sqrt{1691}}{5}$
