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  • Find the length of the chord of the ellipse <img alt="\mathrm{\frac{x^2}{25}+\frac{y^2}{16}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7B25%7D&plus;

Find the length of the chord of the ellipse \mathrm{\frac{x^2}{25}+\frac{y^2}{16}=1} whose middle point is \mathrm{\left(\frac{1}{2}, \frac{2}{5}\right)}.

Option: 1

\frac{7 \sqrt{42}}{5}


Option: 2

\frac{5 \sqrt{42}}{7}


Option: 3

\frac{7 \sqrt{41}}{5}


Option: 4

\frac{5 \sqrt{41}}{7}


Answers (1)

best_answer

Equation of the chord having \left(\frac{1}{2}, \frac{2}{5}\right) as middle point is

\mathrm{\begin{aligned} & \frac{\frac{1}{4}}{25}+\frac{\frac{4}{25}}{16}-1=\frac{\frac{1}{2} x}{25}+\frac{\frac{2}{5} y}{16}-1 \ \ \ \ \ \ \ (S_1 = T) \\ & \Rightarrow 4 x+5 y=4 \Rightarrow 5 y=4(1-x) \end{aligned}}.....[i]

Solving (i) and equation of ellipse, we get

\mathrm{\begin{aligned} & 16 x^2+16(1-x)^2=400 \\ & \Rightarrow x^2+(1-x)^2=25 \Rightarrow x^2-x-12=0 \Rightarrow x=4,-3 \end{aligned}}

\mathrm{\text { When } x=4, y=-\frac{12}{5} \text { and when } x=-3, y=\frac{16}{5}}

\mathrm{\therefore \quad \text { Length of the chord }=\sqrt{(4+3)^2+\left(\frac{-16-12}{5}\right)^2}=\frac{7 \sqrt{41}}{5}}

Posted by

Devendra Khairwa

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