Hi **xwrathbringerx**,

Below, you will find solutions to *Problems 1* and *2*.

Originally Posted by

**xwrathbringerx** 1.The tangent at P on the ellipse x^2/a^2 + y^2/b^2 =1 cuts the x axis at T and the perpendicular PN is drawn to the x-axis. If O is the origin, prove that ON * OT = a^2.

Consider the ellipse

$\displaystyle \mathcal{E}:\ \bigg(\frac{x}{a}\bigg)^{2}+\bigg(\frac{y}{b}\bigg )^{2}=1.$

Let $\displaystyle P=(p,q)\in\mathcal{E}$, then the equation of the tangent line at the point $\displaystyle P$ is given by

$\displaystyle y=-\frac{b^{2}p}{a^{2}q}(x-p)+q$

or eqivalently

$\displaystyle a^{2}qy+b^{2}px=a^{2}b^{2}.$......................(1)

Actually, obtaining the equation of the tangent line is not very simple (just makes use of a simple idea but too much computations) but If you need I can also show it.

To find $\displaystyle T=(t,0)$, we have to plug $\displaystyle T$ into (1), i.e., put $\displaystyle y=0$ and $\displaystyle x=t$.

Therefore, we get

$\displaystyle t=\frac{b^{2}a^{2}}{b^{2}p}=\frac{a^{2}}{p}.$.............................(2)

On the other hand, if $\displaystyle N=(n,0)$ is the projection of $\displaystyle P=(p,q)$, we must have

$\displaystyle n=p.$.........................................(3)

And finally, from (2) and (3), we can find that

$\displaystyle |ON|\times|OT|=n\times t=p\times\frac{a^{2}}{p}=a^{2}$

As a note, the value $\displaystyle \big||OT|-|ON|\big|$ is called the *subtangent length* of the point $\displaystyle P$ (see *Figure 1*). $\displaystyle \rule{0.2cm}{0.2cm}$

Originally Posted by

**xwrathbringerx** 2. Find the equation of the normal l to 9x^2 + 25y^2 = 225 at P(3, 2 2/5). This normal cuts the x-axis at G and N is the foot of the perpendicular drawn from P to the x-axis. Find GN.

You can see from the solution of *Problem 1* that the slope of the tangent line at $\displaystyle P\in\mathcal{E}$ is

$\displaystyle m_{T}=-\frac{b^{2}p}{a^{2}q}.$

Hence, and the point $\displaystyle P$ the normal line must have the slope

$\displaystyle m_{N}=\frac{a^{2}q}{b^{2}p}$

since $\displaystyle m_{T}m_{N}=-1$.

Then, the equation of the normal line is at the point $\displaystyle P\in\mathcal{E}$ is

$\displaystyle y=\frac{a^{2}q}{b^{2}p}(x-p)+q$

or equivalently

$\displaystyle a^{2}qx-b^{2}py=c^{2}pq,$......................(4)

where

$\displaystyle c^{2}:=a^{2}-b^{2}.$

To find $\displaystyle G=(g,0)$, use (4) with $\displaystyle x=g$ and $\displaystyle y=0$, then we get

$\displaystyle g=\frac{c^{2}pq}{a^{2}q}=\frac{c^{2}p}{a^{2}}.$...........................(5)

And the foot $\displaystyle N=(n,0)$ must satisfy

$\displaystyle n=p.$.........................................(6)

So that (5) and (6) yield

$\displaystyle |GN|=\big||OG|-|ON|\big|=\bigg|\frac{c^{2}p}{a^{2}}-p\bigg|=\bigg|\bigg(\frac{c^{2}}{a^{2}}-1\bigg)p\bigg|$

......_$\displaystyle =\bigg|\frac{c^{2}-a^{2}}{a^{2}}p\bigg|$

......_$\displaystyle =\frac{b^{2}}{a^{2}}|p|.$

This value is called the *subnormal length* of the point $\displaystyle P$ (see *Figure 1*).

You may put above $\displaystyle a=5$, $\displaystyle b=3$ and $\displaystyle p=3$ to get your answer. $\displaystyle \rule{0.2cm}{0.2cm}$

**Figure 1**: Subtangent, and subnormal lengths.

Think of the remaining ones with drawing figures,

if you wont be able to succeed I can give a hand again.