Question

The locus of the foot of the perpendicular from the centre of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ on any tangent is given by $(x^2+y^2) = lx^2+my^2$,where

• A. $l= a^2, m = b^2$
• B. $l = -a^2, b = -b^2$
• C. $l = - a^2, m = - b^2$
• D. $l = a^2,b = -b^2$

Answer 1

Answer: (A)

$l= a^2, m = b^2$

Answer 2

Equation of any tangent to the given ellipse is $y= mx\pm\sqrt{a^{2}m^{2}+b^{2}}$ $y-mx =\pm \sqrt{a^{2}m^{2}+b^{2}}\quad...\left(1\right)$ Equation of perpendicular line is $my + x = \lambda$ It passes through the centre $\left(0, 0\right)$ $\therefore \lambda = 0$ $my + x = 0 \quad...\left(2\right)$ On squaring and adding $\left(1\right)$ and $\left(2\right)$, we get $y^{2} +m^{2}x^{2}+m^{2}y^{2}+x^{2} = a^{2}m^{2}+b^{2}$ $\left(1+m^{2}\right)\left(x^{2}+y^{2}\right) = a^{2}m^{2} +b^{2}$ $\Rightarrow \left(1+\frac{x^{2}}{y^{2}}\right)\left(x^{2}+y^{2}\right) = \frac{a^{2}x^{2}}{y^{2}}+b^{2}$ $\quad\left[from \left(2\right)\right]$ $\Rightarrow \left(x^{2}+y^{2}\right)^{2} = a^{2}x^{2}+b^{2}y^{2}$ But $\left(x^{2}+y^{2}\right)^{2} = lx^{2}+my^{2}$ $l=a^{2}, m =b^{2}$