- Home
- Standard 11
- Mathematics
ઉપવલય $\frac{x^2}{25}+\frac{y^2}{16}=1$ ની, $\left(1, \frac{2}{5}\right)$ મધ્યબિંદુ વાળી, જીવાની લંબાઈ .................................છે.
$\frac{\sqrt{1691}}{5}$
$\frac{\sqrt{2009}}{5}$
$\frac{\sqrt{1741}}{5}$
$\frac{\sqrt{1541}}{5}$
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}$
$y=\frac{10-8 x}{5}$ $…….(i)$
$\frac{x^2}{25}+\frac{(10-8 x)^2}{400}=1$ (put in original equation)
$\frac{16 \mathrm{x}^2+100+64 \mathrm{x}^2-160 \mathrm{x}}{400}=1$
$ 4 x^2-8 x-15=0 $
$ x=\frac{8 \pm \sqrt{304}}{8} $
$x_1=\frac{8+\sqrt{304}}{8} ; x_2=\frac{8-\sqrt{304}}{8}$
$\mathrm{x}_1=\frac{8+\sqrt{304}}{8} ; \mathrm{x}_2=\frac{8-\sqrt{304}}{8}$
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 =$\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2}$
$=\sqrt{\frac{4 \times 304}{64}+\frac{4 \times 304}{25}}=\frac{\sqrt{1691}}{5}$
