Question

Find the locus of the middle points of chords of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ which are drawn through the positive end of the minor axis.

• A. $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{y}{b}=0$
• B. $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{y}{b}=0$
• C. $\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{y}{b}=0$
• D. $\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{y}{b}=0$

Answer 1

Answer: (A)

$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{y}{b}=0$

Answer 2

Let the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Assume $a>b$, so the positive end of the minor axis is $(0, b)$. Let $M(h, k)$ be the midpoint of a chord passing through $(0, b)$. Let the endpoints of the chord be $P(x_1, y_1)$ and $Q(x_2, y_2)$. The midpoint $M(h, k)$ has coordinates $h = \frac{x_1+x_2}{2}$ and $k = \frac{y_1+y_2}{2}$. The equation of the chord with midpoint $(h, k)$ is given by $T = S_1$, where $T = \frac{xh}{a^2} + \frac{yk}{b^2}$ and $S_1 = \frac{h^2}{a^2} + \frac{k^2}{b^2}$. So, the equation of the chord is $\frac{xh}{a^2} + \frac{yk}{b^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2}$. Since this chord passes through the point $(0, b)$, we substitute $x=0$ and $y=b$ into the equation of the chord: $\frac{(0)h}{a^2} + \frac{(b)k}{b^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2}$ $\frac{k}{b} = \frac{h^2}{a^2} + \frac{k^2}{b^2}$ Rearranging the terms to find the locus of $(h, k)$: $\frac{h^2}{a^2} + \frac{k^2}{b^2} - \frac{k}{b} = 0$ Replacing $h$ with $x$ and $k$ with $y$ to get the locus equation: $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{y}{b} = 0$

This equation represents an ellipse. It can be rewritten as: $\frac{x^2}{a^2} + \frac{1}{b^2}(y^2 - by) = 0$ $\frac{x^2}{a^2} + \frac{1}{b^2}(y^2 - by + \frac{b^2}{4} - \frac{b^2}{4}) = 0$ $\frac{x^2}{a^2} + \frac{1}{b^2}((y - \frac{b}{2})^2 - \frac{b^2}{4}) = 0$ $\frac{x^2}{a^2} + \frac{(y - \frac{b}{2})^2}{b^2} - \frac{1}{4} = 0$ $\frac{x^2}{a^2} + \frac{(y - \frac{b}{2})^2}{b^2} = \frac{1}{4}$ $\frac{x^2}{a^2/4} + \frac{(y - b/2)^2}{b^2/4} = 1$ This is the equation of an ellipse with center $(0, b/2)$ and semi-axes $a/2$ and $b/2$.