The length of the chord of the ellipse (x^2)/(25)+(y^2)/(16)=1, whose mid point is (1, (2)…

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 :

Accepted Answer

Equation of chord with given middle point.

$\begin{gathered} T={S}_{1} \ \frac{x}{25}+\frac{y}{40}=\frac{1}{25}+\frac{1}{100} \ \frac{8x+5y}{200}=\frac{8+2}{200} \ y=\frac{10-8x}{5}\, \end{gathered}$

$\frac{{x}^{2}}{25}+\frac{(10-8x{)}^{2}}{400}=1$ (put in original equation)

$\begin{gathered} \frac{16{x}^{2}+100+64{x}^{2}-160x}{400}=1 \ 4{x}^{2}-8x-15=0 \ x=\frac{8\pm \sqrt{304}}{8} \ {x}_{1}=\frac{8+\sqrt{304}}{8};{x}_{2}=\frac{8-\sqrt{304}}{8} \end{gathered}$

Similarly, $y=\frac{10-18\pm \sqrt{304}}{5}=\frac{2\pm \sqrt{304}}{5}$

$\begin{gathered} {y}_{1}=\frac{2-\sqrt{304}}{5};{y}_{2}=\frac{2+\sqrt{304}}{5} \ =\sqrt{{({x}_{1}-{x}_{2})}^{2}+{({y}_{1}-{y}_{2})}^{2}} \ =\sqrt{\frac{4	imes 304}{64}+\frac{4	imes 304}{25}}=\frac{\sqrt{1691}}{5} \end{gathered}$

Solutions · 1