ઉપવલય frac{x^2}{25}+frac{y^2}{16}=1 ની, | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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 =$\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2ight)^2+\left(\mathrm{y}_1-\mathrm{y}_2ight)^2}$

$=\sqrt{\frac{4 	imes 304}{64}+\frac{4 	imes 304}{25}}=\frac{\sqrt{1691}}{5}$

ઉપવલય $\frac{x^2}{25}+\frac{y^2}{16}=1$ ની, $\left(1, \frac{2}{5}\right)$ મધ્યબિંદુ વાળી, જીવાની લંબાઈ .................................છે.

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