Hello arze 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 $\displaystyle \frac{b\tan t}{a}$.)

Grandad