The midpoint of the chord of the ellipse x^2 | vedclass

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Question

(B) Let the midpoint of the chord be $M(h, k)$.The equation of the chord of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with midpoint $(h, k)$

Vedclass · Accepted Answer

$(-\frac{1}{5}, \frac{4}{5})$

Vedclass · Answer

(B) Let the midpoint of the chord be $M(h, k)$.The equation of the chord of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with midpoint $(h, k)$ is given by $T = S_1$,where $T = \frac{xh}{a^2} + \frac{yk}{b^2}$ and $S_1 = \frac{h^2}{a^2} + \frac{k^2}{b^2}$.Here,$a^2 = 1$ and $b^2 = 4$. So,the equation is $xh + \frac{yk}{4} = h^2 + \frac{k^2}{4}$.This chord is the same as the line $y = x + 1$,which can be written as $x - y + 1 = 0$,or $x - y = -1$.Comparing the coefficients of $xh + \frac{yk}{4} = h^2 + \frac{k^2}{4}$ and $x - y = -1$:$\frac{h}{1} = \frac{k/4}{-1} = \frac{h^2 + k^2/4}{-1}$.From $\frac{h}{1} = \frac{k}{-4}$,we get $k = -4h$.Substitute $k = -4h$ into $\frac{h}{1} = \frac{h^2 + k^2/4}{-1}$:$-h = h^2 + \frac{(-4h)^2}{4} = h^2 + 4h^2 = 5h^2$.$5h^2 + h = 0 \implies h(5h + 1) = 0$.Since $h$ cannot be $0$ (as the line $y = x + 1$ does not pass through the origin),$h = -1/5$.Then $k = -4(-1/5) = 4/5$.The midpoint is $(-\frac{1}{5}, \frac{4}{5})$.

The midpoint of the chord of the ellipse $x^2 + \frac{y^2}{4} = 1$ formed on the line $y = x + 1$ is

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