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If the tangents are drawn to the ellipse $x^2+2y^2=2$, then the locus of the mid-point of the intercept made by the tangents between the coordinate axis is...

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Differentiating the equation of ellipse, we get $$\frac{dy}{dy}\Bigg|_{(x_1,y_1)}=\frac{-x_1}{2y_1}$$ This is the slope of the tangent at point $(x_1,y_1)$

But I do not know how to proceed from here. Please help!

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Any tangent to the ellipse $x^2+2y^2=2$ is $y=mx+\sqrt{2m^2+1}$ the coordinate of of its intersection with x-axis ia $A(-\sqrt{2m^2+1}/m,0)$ and on yaxis it is $B(0, \sqrt{2m^2+1})$. Let the tangent intercept of AB be $P(h,k)$, the $h=-\frac{\sqrt{2m^2+1}}{2m}, k=\frac{\sqrt{2m^2+1}}{2}$. Eliminiation of $m$ from these equations give: $4h^2k^2=2k^2+h^2$. So the the required locus is $$ 4x^2y^2-2y^2-x^2=0$$

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  • $\begingroup$ Oh! yes I have corrected it now. Thanks. $\endgroup$
    – Z Ahmed
    Oct 6 '20 at 9:53
  • $\begingroup$ Sign of $x^2$ is correct now in the last step. $\endgroup$
    – Z Ahmed
    Oct 6 '20 at 11:13
  • $\begingroup$ Now it looks good. (+1) $\endgroup$ Oct 6 '20 at 11:19
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The tangent: $y=y_1-\frac{x_1}{2y_1}(x-x_1).$

The midpoint: $(\frac{2y_1^2+x_1^2}{2x_1},\frac{2y_1^2+x_1^2}{4y_1})=(\frac1{x_1},\frac1{2y_1}).$

The locus: $(1/x)^2+2(1/(2y))^2/2=2.$

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