If the chord of the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ having $(1,1)$ as its middle point is $\dot{x}+\alpha y=\beta$, then

Question

Aicharya · Accepted Answer

$\frac{x^2}{4}+\frac{y^2}{9}=1$ Equation of chord having $(1,1)$ as its middle point is $\mathrm{T}=\mathrm{S}_1$ $\begin{aligned} & \Rightarrow \frac{x}{4}+\frac{y}{9}-1=\frac{1}{4}+\frac{1}{9}-1 \ & \Rightarrow 9 x+4 y=13 \Rightarrow x+\frac{4}{9} y=\frac{13}{9} \Rightarrow \alpha=\frac{4}{9}, \beta=\frac{13}{9} \ & \Rightarrow \beta-\alpha=1 \Rightarrow \alpha+1=\beta \end{aligned}$

If the chord of the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ having $(1,1)$ as its middle point is…

Solution