If tangents are drawn to the ellipse x^2+2y^2 | vedclass

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(D) The equation of the ellipse is $x^2 + 2y^2 = 2$,which can be written as $\frac{x^2}{2} + \frac{y^2}{1} = 1$.Let the point of tangency be $(x_0, y_

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

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(D) The equation of the ellipse is $x^2 + 2y^2 = 2$,which can be written as $\frac{x^2}{2} + \frac{y^2}{1} = 1$.Let the point of tangency be $(x_0, y_0)$. The equation of the tangent at $(x_0, y_0)$ is $\frac{xx_0}{2} + yy_0 = 1$.The intercepts on the axes are $A = (\frac{2}{x_0}, 0)$ and $B = (0, \frac{1}{y_0})$.Let $(h, k)$ be the midpoint of the intercept $AB$. Then $h = \frac{1}{x_0}$ and $k = \frac{1}{2y_0}$,which implies $x_0 = \frac{1}{h}$ and $y_0 = \frac{1}{2k}$.Since $(x_0, y_0)$ lies on the ellipse,we have $(\frac{1}{h})^2 + 2(\frac{1}{2k})^2 = 2$.This simplifies to $\frac{1}{h^2} + \frac{2}{4k^2} = 2$,or $\frac{1}{h^2} + \frac{1}{2k^2} = 2$.Replacing $(h, k)$ with $(x, y)$,we get $\frac{1}{x^2} + \frac{1}{2y^2} = 2$.

If tangents are drawn to the ellipse $x^2+2y^2=2$,then the locus of the midpoints of the intercepts made by the tangents between the coordinate axes is

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