# Need some clarification

• Apr 27th 2010, 11:05 PM
arze
Need some clarification
AB is a diameter of the ellipse $\displaystyle x^2+9y^2=25$ the eccentric angle at A is $\displaystyle \frac{\pi}{6}$. Find
a) the eccentric angle of B
b) the equations of the tangent at A and B
c) the equation of the conjugate diameter

I found correctly the eccentric angle of B to be $\displaystyle -\frac{5\pi}{6}$

When I did (b) and (c), I used $\displaystyle m_2=-\frac{b^2}{a^2m_1}$ to find the gradient of the tangent and conjugate diameter. $\displaystyle m_1=\tan\frac{\pi}{6}$
But I found that in the answers the gradients are all $\displaystyle \frac{1}{\sqrt{3}}$ which is the gradient of the diameter AB.
$\displaystyle m_2=-\frac{\sqrt{3}}{9}$
I used the equation for tangents $\displaystyle y=mx\pm\sqrt{a^2m^2+b^2}$
Have I misunderstood something?
Thanks
• Apr 28th 2010, 01:47 AM
rooshidavid
Well your method is right but i would suggest you another method that you can use the formula of :
tanθ = ± m1 + m2
1 + m1m2
You will get answer in the form of m. Then you can find the value by elimination method.
• Apr 28th 2010, 01:54 AM
Hello arze
Quote:

Originally Posted by arze
AB is a diameter of the ellipse $\displaystyle x^2+9y^2=25$ the eccentric angle at A is $\displaystyle \frac{\pi}{6}$. Find
a) the eccentric angle of B
b) the equations of the tangent at A and B
c) the equation of the conjugate diameter

I found correctly the eccentric angle of B to be $\displaystyle -\frac{5\pi}{6}$

When I did (b) and (c), I used $\displaystyle m_2=-\frac{b^2}{a^2m_1}$ to find the gradient of the tangent and conjugate diameter. $\displaystyle m_1=\tan\frac{\pi}{6}$ Grandad says: No. See below*
But I found that in the answers the gradients are all $\displaystyle \frac{1}{\sqrt{3}}$ which is the gradient of the diameter AB.
$\displaystyle m_2=-\frac{\sqrt{3}}{9}$
I used the equation for tangents $\displaystyle y=mx\pm\sqrt{a^2m^2+b^2}$
Have I misunderstood something?
Thanks

Are you sure the gradient of the tangent and the conjugate diameter is given as $\displaystyle \frac{1}{\sqrt3}$? I make it $\displaystyle -\frac{1}{\sqrt3}$.

Here's my working.

The equation of the tangent at $\displaystyle (a\cos t, b\sin t)$ is easily shown to be:
$\displaystyle \frac{x\cos t}{a}+\frac{y\sin t}{b}=1$
and the gradient at $\displaystyle (a\cos t, b\sin t)$ is
$\displaystyle -\frac{b\cot t}{a}$
Here $\displaystyle a = 5,\; b = \frac53$ and $\displaystyle t = \frac{\pi}{6}$ at A. So the gradient at A is:
$\displaystyle -\frac{\frac53\cot(\pi/6)}{5}=-\frac{1}{\sqrt3}$
and the tangent at A is
$\displaystyle x\sqrt3+3y=10$
(* Incidentally, the gradient of a diameter isn't $\displaystyle \tan t$. It's $\displaystyle \frac{b\tan t}{a}$.)