# If the normal to the point P on the ellipse intersects the major and minor axis and $G$ and $g$, then find the relation …

If the normal to the point P on the ellipse intersects the major and minor axis and $$G$$ and $$g$$, then find the relation between $$CG$$, $$Cg$$, $$a$$ and $$b$$, where C is centre

The normal at $$P(x_1,y_1)$$ is $$\frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2$$

The intercepts will be $$CG=\frac{x_1(a^2-b^2)}{a^2}$$ $$Cg=\frac{y_1(a^2-b^2)}{b^2}$$

Unfortunately I am not able to eliminate the terms $$x_1$$ and $$y_1$$. How should I do it?

• $(x_1,y_1)$ is on the ellipse, so ... – Blue Apr 7 '20 at 13:33

We know that $$(x_1,y_1)$$ must lie on the ellipse, so, we have: $$\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}=1 \leftrightarrow y_1^2=b^2\left(1-\frac{x_1^2}{a^2}\right)$$ Substituing, we obtain: $$Cg^2=\frac{b^2\left(1-\frac{x_1^2}{a^2}\right)(a^2-b^2)^2}{b^4}$$ While $$CG^2$$ is: $$CG^2=\frac{x_1^2(a^2-b^2)^2}{a^4}\leftrightarrow x_1^2=\frac{CG^2\cdot a^4}{(a^2-b^2)^2}$$ Substituing again, you arrive at: $$Cg^2=\frac{b^2\left(1-\frac{CG^2\cdot a^4}{a^2(a^2-b^2)^2}\right)(a^2-b^2)^2}{b^4} \rightarrow Cg^2=\frac{(a^2-b^2)^2-CG^2\cdot a^2}{b^2}$$ From here, it's very simple. In fact: $$Cg=\sqrt{Cg^2}=\sqrt{\frac{(a^2-b^2)^2-CG^2\cdot a^2}{b^2}}=\frac{\sqrt{(a^2-b^2)^2-CG^2\cdot a^2}}{b}$$

Note that, as @Blue suggested, you can rewrite the relation for $$Cg^2$$ as: $$\frac{Cg^2}{a^2} + \frac{CG^2}{b^2} =\frac{ (a^2-b^2)^2}{a^2b^2}$$ This follows from here: $$Cg^2=\frac{(a^2-b^2)^2-CG^2\cdot a^2}{b^2} \leftrightarrow Cg^2b^2-CGa^2=(a^2-b^2)^2 \leftrightarrow \frac{Cg^2}{a^2} + \frac{CG^2}{b^2} =\frac{ (a^2-b^2)^2}{a^2b^2}$$

• +1. I probably would have manipulated your arrived-at relation into $$\frac{Cg^2}{a^2} + \frac{CG^2}{b^2} =\frac{ (a^2-b^2)^2}{a^2b^2}$$ to maximize symmetry. – Blue Apr 7 '20 at 13:50
• @Blue: I will add this in the answer. – Matteo Apr 7 '20 at 13:54
• Done. Thanks!!! – Aditya Apr 7 '20 at 17:29
• @Aditya: none, thanks. I like a lot problems on conic section, especially ellipse, so I really like this questions. – Matteo Apr 7 '20 at 17:30

In the same vein as Matteo’s answer but with, I think, much less work: From your calculations we have $$x_1 = {a^2\over a^2-b^2}{CG} \\ y_1 = {b^2\over a^2-b^2}{Cg}.$$ The point $$(x_1,y_1)$$ lies on the ellipse, therefore $${x_1^2\over a^2} + {y_1^2\over b^2} = \frac1{a^2}\left({a^2\over a^2-b^2}{CG}\right)^2 + \frac1{b^2}\left({b^2\over a^2-b^2}{Cg}\right)^2 = 1$$ and after some straightforward rearrangement, we end up with $${{Cg}^2\over a^2}+{{CG}^2\over b^2} = {(a^2-b^2)^2\over a^2b^2}.$$

WLOG any point on the ellipse $$P(a\cos t,b\sin t)$$

the equation of the normal $$ax\sec t-by\csc t=a^2-b^2$$

$$G: \dfrac{(a^2-b^2)\cos t}a,0$$

$$g: 0, -\dfrac{(a^2-b^2)\sin t}b$$

$$a\cdot CG=|(a^2-b^2)\cos t|,|\cos t|=?$$

Similarly

$$b\cdot CG=|(a^2-b^2\sin t|,|\sin t|=?$$

Eliminate $$t$$ using $\cos^2t+\sin^2t=1$\$