Question

Find the locus of the middle point of the intercept of the tangents drawn from an external point to the ellipse $x^2+2y^2=2$ between the coordinate axes?

Answer 1

Consider the equation of the ellipse.

$\Rightarrow$$x^2+2y^2=2$

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

$\therefore$ $a=\sqrt{2}$ and $b=1$

Consider the diagram shown above. Let AB be the tangent on the ellipse. Let $(h,k)$ be the mid-point.

Let $P$ be any point on the ellipse. Here,

$\Rightarrow P(a\cos \theta ,b\sin \theta )$

Therefore, equation of the tangent at point $P$ be,

$\Rightarrow$$\frac{x{x}_1}{2}+y{y}_1=1$

$\Rightarrow$$\frac{x\sqrt{2}\cos \theta }{2}+y\sin \theta =1$

$\Rightarrow$$\frac{x\cos \theta }{\sqrt{2}}+y\sin \theta =1$

Therefore,

$\Rightarrow A(\sqrt{2}\sec \theta ,0)$

$\Rightarrow B(0,cosec\theta )$

Therefore,

$\Rightarrow$$h=\frac{\sqrt{2}\sec \theta +0}{2},k=\frac{0+cosec\theta }{2}$

Therefore,

$\Rightarrow$$\cos \theta =\frac{1}{\sqrt{2h}},\sin \theta =\frac{1}{2k}$

Now, we know that,

$\Rightarrow$${\cos }^2\theta +{\sin }^2\theta =1$

$\Rightarrow$$\frac{1}{2h^2}+\frac{1}{4k^2}=1$

Put $h=x,k=y$.

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

Hence, this is the required locus.