Question

Find the middle point of the chord intercepted on the line $2x-y+3=0$ by the ellipse $\frac{x^2}{10}+\frac{y^2}{6}=1$.

• A. $\left(\begin{array}{c}\frac{-30}{23},\frac{9}{23}\end{array}\right)$
• B. $\left(\begin{array}{c}\frac{30}{23},\frac{9}{23}\end{array}\right)$
• C. $\left(\begin{array}{c}\frac{30}{24},\frac{9}{24}\end{array}\right)$
• D. None of these

Answer 1

The correct option is A $\left(\begin{array}{c}\frac{-30}{23},\frac{9}{23}\end{array}\right)$

Given equation of the chord is $2x-y+3=0$, ellipse equation is $\frac{x^2}{10}+\frac{y^2}{6}=1---\left(i\right)$.

The equation of the chord of the ellipse $S=0$ having $P(x_1,y_1)$ as it's midpoint is $S_1=S_{11}$

$\frac{x{x}_1}{10}+\frac{y{y}_1}{6}=\frac{x_1^2}{10}+\frac{y_1^2}{6}---\left(ii\right)$

comparing $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow \frac{2}{\frac{x_1}{10}}=\frac{-1}{\frac{y_1}{6}}=\frac{-3}{\frac{x_1^2}{10}+\frac{y_1^2}{6}}$

$\Rightarrow \frac{20}{x_1}=-\frac{6}{y_1}\Rightarrow -\frac{10}{3}=\frac{x_1}{y_1}\to \left(1\right)$

substitute $\left(1\right)$ in $2x-y+3=0$ , we get

$y_1=\frac{9}{23}\&x_1=\frac{-30}{23}$

So, the mid point is $(\frac{-30}{23},\frac{9}{23})$