Question

Find the length of the chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, whose middle point is $(\frac{1}{2},\frac{2}{5})$.

Answer 1

Chord of ellipse with mid-point $(x_1,y_1)$ is $T=S_1$

$\therefore \frac{x}{50}+\frac{y}{40}-1=\frac{1}{25\times 4}+\frac{4}{25\times 16}-1$

$\therefore 4x+5y=4$

This line when solved with the ellipse gives the points of intersection.

$\therefore \frac{x^2}{25}+\frac{{\left(\frac{4(1-x)}{5}\right)}^2}{16}=1$

$\therefore \frac{x^2}{25}+\frac{16(1-x{)}^2}{16\times 25}=1$

$\therefore \frac{x^2+(1-x{)}^2}{25}=1$

$\therefore 2x^2-2x-24=0$

$\therefore x=-3$ or $x=4$

Hence, the points of intersection are $(-3,\frac{16}{5})$ and $(4,-\frac{12}{5})$.

$\therefore$ length of chord $=\sqrt{(-3-4{)}^2+{(\frac{16}{5}+\frac{12}{5})}^2}$

$=\sqrt{(7{)}^2+{\left(\frac{4\times 7}{5}\right)}^2}$

$=7\sqrt{1+\frac{16}{25}}$

$=\frac{7}{5}\sqrt{41}$

This is the required answer.