# Thread: locus and ellipse problems

1. ## locus and ellipse problems

Hi,

You can probably tell by now that me and conic section/locus questions don't get on.

This time we are considering an ellipse with equation

$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and focii at S and S'.

Consider an arbitrary point P on the ellipse, I need to show that the line SP and the perpendicular from the center O to the tangent at P meet at G, and that the locus of G is a circle centre S with radius a.

Well the gradient of the tangent at a point P(a cos(t), b sin(t)) is $\displaystyle \frac{-b \cos t}{a\sin t}$ so the gradient of the perpendicular to the tangent is $\displaystyle \frac{a \sin t}{b \cos t}$, this passes through the origin and so the equation for this line is $\displaystyle y= \frac{a \sin t}{b \cos t} x$

Now the gradient of the line SP is $\displaystyle \frac{\Delta y}{\Delta x} = \frac{b \sin t}{a \cos t + ae}$. This passes through the point (-ae, 0) and so the equation of the line SP is

$\displaystyle y=\frac{b \sin t}{a \cos t + ae}(x+ae)$

Presumably I must solve these equations simultaneously and then eliminate t to find the required equation of the circle. Since there is a lone y on each equation I tried setting them equal but got a horrific mess with the algebra so I think there's a trick I might be missing

Any help on how to solve these equations simultaneously (or even if my method is correct!) would be very much appreciated thank you.

2. So you have the equations $\displaystyle y= \frac{a \sin t}{b \cos t} x$ and $\displaystyle y=\frac{b \sin t}{a \cos t + ae}(x+ae)$, and you want to eliminate t between them.

Write the first equation as $\displaystyle \tan t = \frac{by}{ax}$ and think of $\displaystyle by$ and $\displaystyle ax$ as sides in a right-angled triangle. Pythagoras' theorem tells you that $\displaystyle \cos t = ax/D$ and $\displaystyle \sin t = by/D$, where $\displaystyle D = \sqrt{a^2x^2+b^2y^2}$. Substitute those values for cos(t) and sin(t) into the second equation, do a bit of cancelletion, use the fact that $\displaystyle 1-e^2 = b^2/a^2$, and it simplifies to $\displaystyle D = -aex+b^2$. Now square both sides to get $\displaystyle a^2x^2 + b^2y^2 = a^2e^2x^2 - 2b^2aex + b^4$, and a bit more simplification reduces that to $\displaystyle (x+ae)^2 + y^2 = a^2$.

3. Brilliant! Thanks