Find the normals to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ which are farthest from the centre.
My approach
Let the point be $(x_1 ,y_1)$ and the tangent equation is $\frac{xx_1}{9}+\frac{yy_1}{4}=1$
Let the point be $(x_1 ,y_1)$ and the normal equation be $\frac{xy_1}{4}-\frac{yx_1}{9}=\frac{x_1y_1}{4}-\frac{x_1y_1}{9}=\frac{5x_1y_1}{36}$
The equation reduced to $9xy_1-4yx_1=5x_1y_1$, not able to proceed from here
Now, let $x_1=a$ and $y_1=b$.
Thus, the distance from the normal to the origin it's $$\frac{5|ab|}{\sqrt{16a^2+81b^2}}$$ or after homogenization it's $$\frac{5|ab|}{\sqrt{(16a^2+81b^2)\left(\frac{a^2}{9}+\frac{b^2}{4}\right)}}.$$ We'll prove that the maximal value of the last expression is $2$.
Indeed, we need to prove that $$\frac{5|ab|}{\sqrt{(16a^2+81b^2)\left(\frac{a^2}{9}+\frac{b^2}{4}\right)}}\leq2$$ or $$(16a^2+81b^2)(4a^2+9b^2)\geq900a^2b^2$$ or $$(8a^2-27b^2)^2\geq0,$$ which is obvious.
An equality occurs for $8a^2-27b^2=0$, which says that $2$ is a maximal value and it'a remains to solve the following system: $$8a^2=27b^2,$$ $$\frac{a^2}{9}+\frac{b^2}{4}=1$$ and to write an equations of the normals.