# The locus of the mid-point of the intercept made by the tangents between the coordinate axis is...

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...

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 $$(x_1,y_1)$$

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 $$4x^2y^2-2y^2-x^2=0$$
• Sign of $x^2$ is correct now in the last step. Oct 6 '20 at 11:13
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: $$(1/x)^2+2(1/(2y))^2/2=2.$$