Question

The equation to the locus of the middle points of the portion of the tangent to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ included between the co-ordinate axes is the curve.

• A. $9x^2+16y^2=4x^2{y}^2$
• B. $16x^2+9y^2=4x^2{y}^2$
• C. $3x^2+4y^2=4x^2{y}^2$
• D. $9x^2+16y^2=x^2{y}^2$

Answer 1

The correct option is A $9x^2+16y^2=4x^2{y}^2$

Let any tangent of ellipse is

$\frac{x\cos \theta }{4}+\frac{y\sin \theta }{3}=1$

Let it meets axes at $A(\frac{4}{\cos \theta },0)\&B(0,\frac{3}{\sin \theta })$

Let midpoint of $AB$ is $(h,k)$ then

$2h=\frac{4}{\cos \theta },2k=\frac{3}{\sin \theta }$

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

$\therefore \frac{16}{4h^2}+\frac{9}{4k^2}=1$

$\Rightarrow 16k^2+9h^2=4h^2{k}^2$

Hence, locus is $16y^2+9x^2=4x^2{y}^2$