Question

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

Answer 1

Dear student

$$\begin{gathered} \frac{x^2}{25}+\frac{y^2}{16}=1 \\ Middlepointis\left(\frac{1}{2},\frac{2}{5}\right) \\ Let\text{'}T\text{'}betheeqofthechord. \\ Let\text{'}S\text{'}betheeqoftheellipse \\ Thenthechordhavingitsmiddlepointas\left(x_1,y_1\right)isgivenby \\ T=S_1 \\ Here,T=\frac{x{x}_1}{25}+\frac{y{y}_1}{16}-1 \\ {S}_1=\frac{{x_1}^2}{25}+\frac{{y_1}^2}{16}-1 \\ Put,x_1=\frac{1}{2},y_1=\frac{2}{5} \\ Theeqofthechordis: \\ \frac{x}{50}+\frac{y}{40}=\frac{x}{100}+\frac{y}{100} \\ \frac{4x+5y}{200}=\frac{2}{100} \\ 4x+5y=4 \\ 5y=4-4x \\ y=\frac{4}{5}\left(1-x\right) \\ Puttingineqofellipse \\ \frac{x^2}{25}+\frac{y^2}{16}=1 \\ \frac{x^2}{25}+\frac{{\left({\displaystyle \frac{4}{5}}\left(1-x\right)\right)}^2}{16}=1 \\ \frac{x^2}{25}+\frac{{\left(1-x\right)}^2}{25}=1 \\ \frac{x^2+1+x^2-2x}{25}=1 \\ 2x^2-2x+1=25 \\ 2x^2-2x-24=0 \\ {x}^2-x-12=0 \\ {x}^2-4x+3x-12=0 \\ \left(x-4\right)\left(x+3\right)=0 \\ Hence,x=4orx=-3 \\ Forx=4 \\ y=\frac{4}{5}\left(1-x\right)=\frac{4}{5}\left(1-4\right)=\frac{-12}{5} \\ Forx=-3 \\ y=\frac{4}{5}\left(1-x\right)=\frac{4}{5}\left(1+3\right)=\frac{16}{5} \\ Henceendpointsare \\ A\left(4,\frac{-1}{5}\right) \\ B\left(3,\frac{16}{5}\right) \\ AB=\sqrt{{\left(3+4\right)}^2+{\left(\frac{16}{5}+\frac{1}{5}\right)}^2}=\sqrt{49+\frac{784}{25}}=\frac{1}{5}\sqrt{2009}units \end{gathered}$$
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