The locus of the midpoints of the segments of | vedclass

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(D) Let the midpoint of the segment $PQ$ intercepted between the axes be $R(x_1, y_1)$.Then the coordinates of $P$ and $Q$ are $(2x_1, 0)$ and $(0, 2y

Vedclass · Accepted Answer

$\frac{a^2}{x^2} + \frac{b^2}{y^2} = 4$

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(D) Let the midpoint of the segment $PQ$ intercepted between the axes be $R(x_1, y_1)$.Then the coordinates of $P$ and $Q$ are $(2x_1, 0)$ and $(0, 2y_1)$ respectively.The equation of the line $PQ$ is $\frac{x}{2x_1} + \frac{y}{2y_1} = 1$,which can be written as $y = -\left(\frac{y_1}{x_1}\right)x + 2y_1$.If this line is tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,it must satisfy the condition $c^2 = a^2m^2 + b^2$,where $m = -\frac{y_1}{x_1}$ and $c = 2y_1$.Substituting these values,we get $(2y_1)^2 = a^2(-\frac{y_1}{x_1})^2 + b^2$.$4y_1^2 = \frac{a^2y_1^2}{x_1^2} + b^2$.Dividing both sides by $y_1^2$,we get $4 = \frac{a^2}{x_1^2} + \frac{b^2}{y_1^2}$.Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $\frac{a^2}{x^2} + \frac{b^2}{y^2} = 4$.

The locus of the midpoints of the segments of the tangents to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ intercepted between the coordinate axes is:

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