Chords of an ellipse frac{x^2}{a^2} + frac{y^ | vedclass

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(C) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The positive end of the minor axis is $P(0, b)$.Let $(h, k)$ be the midpoint of a chor

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an ellipse

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(C) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The positive end of the minor axis is $P(0, b)$.Let $(h, k)$ be the midpoint of a chord passing through $(0, b)$ and some point $(x_1, y_1)$ on the ellipse.Then,$h = \frac{x_1 + 0}{2} \Rightarrow x_1 = 2h$ and $k = \frac{y_1 + b}{2} \Rightarrow y_1 = 2k - b$.Since $(x_1, y_1)$ lies on the ellipse,we have $\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} = 1$.Substituting the values of $x_1$ and $y_1$,we get $\frac{(2h)^2}{a^2} + \frac{(2k - b)^2}{b^2} = 1$.This simplifies to $\frac{4h^2}{a^2} + \frac{4(k - b/2)^2}{b^2} = 1$,which can be rewritten as $\frac{h^2}{(a/2)^2} + \frac{(k - b/2)^2}{(b/2)^2} = 1$.Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{x^2}{(a/2)^2} + \frac{(y - b/2)^2}{(b/2)^2} = 1$,which represents an ellipse.

Chords of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are drawn through the positive end of the minor axis $(0, b)$. The locus of their midpoints lies on:

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