# Tangent of ellipse

• Apr 27th 2010, 07:25 PM
arze
Tangent of ellipse
The tangent at $\displaystyle P(4\cos\theta, 3\sin\theta)$ on the ellipse $\displaystyle 9x^2+16y^2=144$ meets the tangent at the positive end of the major axis at Q and the positive end of the minor axis at R. Find
a) the ratio of the lengths PQ and PR
b) the parametric equations of the midpoint of QR.

The equation of the tangent at the point P is $\displaystyle bx\cos\theta+ay\sin\theta=ab$
plugging the values i got
$\displaystyle 3x\cos\theta+4y\sin\theta=12$
then i found the point where it met the axes
$\displaystyle P(4\sec\theta,0)$
and $\displaystyle Q(0,3\csc\theta)$
The i found PQ and PR
$\displaystyle PQ=\sqrt{(4\cos\theta-4\sec\theta)^2+(3\sin\theta)^2}$
$\displaystyle =\sin\theta\sqrt{16\tan^2\theta+9}$
and $\displaystyle PR=\sqrt{(4\cos\theta)^2+(3\sin\theta-3\csc\theta)^2}$
$\displaystyle =\frac{\cos\theta}{\tan\theta}\sqrt{16\tan^2\theta +9}$
then the ratio would be $\displaystyle \sin\theta:\frac{\cos\theta}{\tan\theta}$
but the answer is $\displaystyle \cos\theta(1-\cos\theta):\sin\theta(1-\sin\theta)$
Thanks for any help!
• Apr 28th 2010, 12:00 AM
Opalg
Quote:

Originally Posted by arze
The tangent at $\displaystyle P(4\cos\theta, 3\sin\theta)$ on the ellipse $\displaystyle 9x^2+16y^2=144$ meets the tangent at the positive end of the major axis at Q and the positive end of the minor axis at R. Find
a) the ratio of the lengths PQ and PR
b) the parametric equations of the midpoint of QR.

The equation of the tangent at the point P is $\displaystyle bx\cos\theta+ay\sin\theta=ab$
plugging the values i got
$\displaystyle 3x\cos\theta+4y\sin\theta=12$
then i found the point where it met the axes
$\displaystyle P(4\sec\theta,0)$
and $\displaystyle Q(0,3\csc\theta)$
The i found PQ and PR
$\displaystyle PQ=\sqrt{(4\cos\theta-4\sec\theta)^2+(3\sin\theta)^2}$
$\displaystyle =\sin\theta\sqrt{16\tan^2\theta+9}$
and $\displaystyle PR=\sqrt{(4\cos\theta)^2+(3\sin\theta-3\csc\theta)^2}$
$\displaystyle =\frac{\cos\theta}{\tan\theta}\sqrt{16\tan^2\theta +9}$
then the ratio would be $\displaystyle \sin\theta:\frac{\cos\theta}{\tan\theta}$
but the answer is $\displaystyle \cos\theta(1-\cos\theta):\sin\theta(1-\sin\theta)$

You have taken P and Q to be the points where the tangent meets the coordinate axes. But the question says that P and Q are the points where the tangent meets the axes of the ellipse, namely the lines x=4 and y=3.
• Apr 28th 2010, 12:35 AM
arze
sorry, i thought the major and minor axes where the center was at the origin would be the x and y axes?
• Apr 28th 2010, 12:46 AM
Opalg
Quote:

Originally Posted by arze
sorry, i thought the major and minor axes where the center was at the origin would be the x and y axes?

My mistake, I should have said what the question says, which is:
The tangent at $\displaystyle P(4\cos\theta, 3\sin\theta)$ on the ellipse $\displaystyle 9x^2+16y^2=144$ meets the tangent at the positive end of the major axis at Q and the positive end of the minor axis at R.
The last part of that sentence is a bit misleading. It must mean that the tangent at P meets the tangent at the positive end of the minor axis at R. You need to take Q and R to be the points where the tangent at P meets the lines x=4 and y=3.
• Apr 28th 2010, 12:55 AM
arze
Ok i understand know, haha my mistake. Thanks!