The equation of the chord of the ellipse 2x² + 5y² = 20 bisected at the point (2,1) is :

a). 4x + 5y + 13 = 0
​b). 4x + 5y = 13
c). 5x+4y+13 = 0
d). 5x+4y = 13

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Equation} \mathrm{of} \mathrm{chord} \mathrm{of} \mathrm{ellipse} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \mathrm{having} \mathrm{mid} \mathrm{point} \left({\mathrm{x}}_1,{\mathrm{y}}_1\right) \mathrm{is} \\ \frac{{\mathrm{xx}}_1}{{\mathrm{a}}^2}+\frac{{\mathrm{yy}}_1}{{\mathrm{b}}^2}=\left(\frac{{\mathrm{x}}_1}{{\mathrm{a}}^2}+\frac{{{\mathrm{y}}_1}^2}{{\mathrm{b}}^2}\right) \\ 2{\mathrm{x}}^2+5{\mathrm{y}}^2=20 \\ \Rightarrow \frac{{\mathrm{x}}^2}{10}+\frac{{\mathrm{y}}^2}{4}=1 \\ \mathrm{Hence} \mathrm{equation} \mathrm{of} \mathrm{chord} \mathrm{will} \mathrm{be} \\ \frac{{\mathrm{xx}}_1}{10}+\frac{{\mathrm{yy}}_1}{4}=\left(\frac{{{\mathrm{x}}_1}^2}{10}+\frac{{{\mathrm{y}}_1}^2}{4}\right) \\ \mathrm{Given} \mathrm{that} \left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\left(2,1\right) \\ \frac{2\mathrm{x}}{10}+\frac{\mathrm{y}}{4}=\left(\frac{4}{10}+\frac{1}{4}\right) \\ \Rightarrow \frac{\mathrm{x}}{5}+\frac{\mathrm{y}}{4}=\left(\frac{2}{5}+\frac{1}{4}\right) \\ \Rightarrow \frac{4\mathrm{x}+5\mathrm{y}}{20}=\frac{8+5}{20} \\ \mathbf{\Rightarrow }\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{13}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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