Question

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

Answer 1

$(-\frac{30}{23}, \frac{9}{23})$

Answer 2

Let the middle point of the chord be $(x_1, y_1)$. The equation of the chord with middle point $(x_1, y_1)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $T=S_1$, which is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2}$.

In this case, $a^2 = 10$, $b^2 = 6$. The equation of the chord is $\frac{xx_1}{10} + \frac{yy_1}{6} = \frac{x_1^2}{10} + \frac{y_1^2}{6}$.

This chord is the same as the given line $2x-y+3=0$.

Comparing the coefficients of $x$, $y$, and the constant terms, we have $\frac{x_1/10}{2} = \frac{y_1/6}{-1} = \frac{-(x_1^2/10 + y_1^2/6)}{3}$.

Let the common ratio be $k$. Then $x_1 = 20k$ and $y_1 = -6k$.

Also, $3k = -(x_1^2/10 + y_1^2/6)$. Substituting the expressions for $x_1$ and $y_1$ in terms of $k$, we get $3k = -(\frac{(20k)^2}{10} + \frac{(-6k)^2}{6}) = -(40k^2 + 6k^2) = -46k^2$.

So $46k^2 + 3k = 0$, which gives $k(46k+3) = 0$.

Since the line does not pass through the center, $k \neq 0$. Thus $46k+3=0$, which means $k = -\frac{3}{46}$.

Then $x_1 = 20k = 20(-\frac{3}{46}) = -\frac{30}{23}$ and $y_1 = -6k = -6(-\frac{3}{46}) = \frac{9}{23}$.

The middle point is $(-\frac{30}{23}, \frac{9}{23})$.