Question

The locus of the mid-point of the portion of the tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ intercepted between the axes is

• A. $a^2{y}^2+b^2{x}^2=4x^2{y}^2$
• B. $a^2{x}^2+b^2{y}^2=4x^2{y}^2$
• C. $x^2+y^2=a^2$
• D. $x^2+y^2=b^2$

Answer 1

The correct option is A $a^2{y}^2+b^2{x}^2=4x^2{y}^2$

Any tangent to the ellipse is $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$

It meets axes at

$A(\frac{a}{\cos \theta },0)B\left(\frac{b}{\sin \theta }\right)$

If $(b,k)$ be the midpoint of $AB$ then $2h=\frac{a}{\cos \theta },2k=\frac{b}{\sin \theta }$

$\therefore \cos \theta =\frac{a}{2h};\sin \theta =\frac{b}{2k}$

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

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

$a^2{y}^2+b^2{x}^2=4x^2{y}^2$