Consider a tangent to the ellipse frac{x^{2}} | vedclass

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(D) Let the point on the ellipse be $P(x_0, y_0)$. The equation of the tangent at $P$ is $\frac{x x_0}{2} + y y_0 = 1$.The intercepts of this tangent

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$\frac{1}{2 x^{2}}+\frac{1}{4 y^{2}}=1$

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(D) Let the point on the ellipse be $P(x_0, y_0)$. The equation of the tangent at $P$ is $\frac{x x_0}{2} + y y_0 = 1$.The intercepts of this tangent on the $x$-axis and $y$-axis are found by setting $y=0$ and $x=0$ respectively.For $y=0$, $x = \frac{2}{x_0}$. For $x=0$, $y = \frac{1}{y_0}$.Let the midpoint of the intercepted portion be $(h, k)$. Then $h = \frac{1}{x_0}$ and $k = \frac{1}{2 y_0}$.This gives $x_0 = \frac{1}{h}$ and $y_0 = \frac{1}{2 k}$.Since $(x_0, y_0)$ lies on the ellipse $\frac{x_0^2}{2} + y_0^2 = 1$, we substitute the values:$\frac{(1/h)^2}{2} + (1/2k)^2 = 1$$\frac{1}{2h^2} + \frac{1}{4k^2} = 1$.Replacing $(h, k)$ with $(x, y)$, the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

Consider a tangent to the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{1}=1$ at any point. The locus of the midpoint of the portion intercepted between the axes is

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