Question

In the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, write down the equations to the diameters which are conjugate to the diameters whose equations are $x-y=0,x+y=0,y=\frac{a}{b}x$, and $y=\frac{b}{a}x$.

Answer 1

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1x-y=0y=xm=1$

Let $m^{\prime }$ be the slope of conjugate diameter

$m{m}^{\prime }=-\frac{b^2}{a^2}\left(1\right)m^{\prime }=-\frac{b^2}{a^2}{m}^{\prime }=-\frac{b^2}{a^2}$

Equation of conjugate diameter $y=m^{\prime }x$

$y=-\frac{b^2}{a^2}x{a}^{2}y+b^{2}x=0$

$-------------------------------------$

$x+y=0y=-xm=-1m{m}^{\prime }=-\frac{b^2}{a^2}(-1)m^{\prime }=-\frac{b^2}{a^2}{m}^{\prime }=\frac{b^2}{a^2}y=m^{\prime }xy=\frac{b^2}{a^2}x{a}^{2}y-b^{2}x=0$

$----------------------------------------$

$y=\frac{a}{b}xm=\frac{a}{b}m{m}^{\prime }=-\frac{b^2}{a^2}\frac{a}{b}{m}^{\prime }=-\frac{b^2}{a^2}{m}^{\prime }=-\frac{b^3}{a^3}y=m^{\prime }xy=-\frac{b^3}{a^3}x{a}^{3}y+b^{3}x=0$

$------------------------------------$

$y=\frac{b}{a}xm=\frac{b}{a}m{m}^{\prime }=-\frac{b^2}{a^2}\frac{b}{a}{m}^{\prime }=-\frac{b^2}{a^2}{m}^{\prime }=-\frac{b}{a}y=m^{\prime }xy=-\frac{b}{a}xay+bx=0$