The mid point of the chord 2x + 5y = 12 of the ellipse 4x² + 5y² = 20 is
 

$$\begin{gathered} Given ellipse S\equiv 4x^2+5y^2-20=0 \\ Let P(x_1,y_1) be mid point of chord \\ So chord equation is S_1=S_{11} \\ \Rightarrow 4x{x}_1+5y{y}_1=4{{x_1}^2}_{ }+5{y_1}^2\to \left(1\right) \\ Given chord 2x+5y=12\to \left(2\right) \\ Since \left(1\right) and \left(2\right) represents same line then \\ \frac{4x_1}{2}=\frac{5y_1}{5}=\frac{4{{x_1}^2}_{ }+5{y_1}^2}{12} \\ \Rightarrow y_1=2x_1 and 4{{x_1}^2}_{ }+5{y_1}^2=12y_1 \\ \Rightarrow 24{x_1}^2=24x_1 \\ \Rightarrow x_1=1 and y_1=2 \\ So mid point=(1,2) \end{gathered}$$