Find the pedal equation of the ellipse $\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$
My Attempt:
Given equation of ellipse is
$$\frac {x^2}{a^2} + \frac {y^2}{b^2}=1$$
Differentiating both sides wrt $x$
$$\frac {2x}{a^2} + \frac {2y}{b^2} \cdot \frac {dy}{dx} = 0$$
$$\frac {dy}{dx}=-\frac {b^2x}{a^2y}$$
Equation of tangent at point $(x,y)$ is
$$Y-y=-\frac {b^2x}{a^2y} (X-x)$$
$$b^2x X+ a^2y Y-a^2y^2-b^2x^2=0$$
Now, length of perpendicular from $(0,0)$ to the tangent is
$$p=\frac {-(a^2y^2+b^2x^2)}{\sqrt {b^4x^2+a^4y^2}}$$
How to solve further?
Best Answer
The equation of the tangent to given ellipse at the point $~(X,Y),~$$~\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1~,\tag1$ is$~\dfrac {xX}{a^2} + \dfrac {yX}{b^2} = 1~.\tag2$
Compared the equation of the tangent of the ellipse with $~AX + BY + C = 0 ~,$ we have $~A=\frac {x}{a^2},~B=\frac {y}{b^2}~$ and $~C=-1~.$
Now perpendicular distance of the tangent of the ellipse form $~(0,0)~$ is $$p=\left|\frac {C}{\sqrt{A^2+B^2}}\right|=\left|\frac {-1}{\sqrt{\left(\frac {x}{a^2}\right)^2+\left(\frac {y}{b^2}\right)^2}}\right|\implies \frac 1{p^2}=\frac {x^2}{a^4} + \frac {y^2}{b^4}~.\tag3$$ We know that $$~r^2=x^2+y^2\implies r^2-b^2=x^2+(y^2-b^2)$$ $$\implies\frac{r^2-b^2}{a^2-b^2}=\frac{x^2+(y^2-b^2)}{a^2-b^2}=\frac{x^2-\frac{b^2}{a^2}x^2}{a^2-b^2}=\dfrac {x^2}{a^2}~.\qquad[\text{using equation $(1)$}]$$ Hence $~\dfrac {x^2}{a^2}=\dfrac{r^2-b^2}{a^2-b^2}~$.
Similarly, $~\dfrac {y^2}{b^2}=\dfrac{a^2-r^2}{a^2-b^2}~$.
Putting these two values in equation $(3)$, we have $$\frac 1{p^2}=\frac {1}{a^2}\left(\frac{r^2-b^2}{a^2-b^2}\right) + \frac {1}{b^2}\left(\frac{a^2-r^2}{a^2-b^2}\right)$$ $$\implies \frac {a^2b^2}{p^2}=\frac {(r^2b^2-b^4)+(a^4-a^2r^2)}{a^2-b^2}=(a^2+b^2-r^2)$$ Hence the pedal equation of the ellipse $(1)$ is $$ \frac {a^2b^2}{p^2}=(a^2+b^2-r^2)~.$$