# co-ordinate geometry

• March 24th 2010, 05:37 PM
jiboom
co-ordinate geometry
it's late and i cant get these 2 question to fall out! help please :)

I have point $p(3\sqrt2 cos \theta,3sin \theta)$ on ellipse

$
\frac{x^2}{18}+\frac{y^2}{9}=1
$

I have the tangent and normal to this ellipse meet Y-axis at T and N. Then i found the centre of the circle through T,N,P to be

$
X=(0,\frac{3}{2}[\frac{1-sin^2\theta}{sin\theta}]
$

The last part asks for the equation of the locus of Q on PX produced such that X is the midpoint of PQ

so ive let Q=(x,y) then

$
\frac{x+3\sqrt(2)}{2}=0
$

$
\frac{y+3sin\theta}{2}=\frac{3}{2}[\frac{1-sin^2\theta}{sin\theta}]
$

so $Q=(-3\sqrt(2),3[\frac{1-2sin^2\theta}{sin\theta}])$

but i cant find the locus. The book says

$
\frac{2x^2}{9}+\frac{9}{16y^2}=1
$

i cant get this with my Q. Maybe im making a mistake or my Q is wrong or the book is wrong......

2)

Also having trouble with this:

the tangent at $P(a cos\theta,bsin\theta)$ to

ellipse $b^2x^2+a^2y^2=a^2b^2$

cuts the Y-axis at Q. The normal at P is parallell to the line joining Q to one focus S'. If S is the other focus, show that PS is parallel to the y-axis.

found Q to be $(0,\frac{b}{sin\theta})$

focus $S'=(ae,0)$ $S=(-ae,0)$ with $a^2-b^2=a^2e^2$

grad $QS=\frac{\frac{b}{sin\theta}}{-ae}=\frac{-b}{aesin\theta}$

so

$
\frac{asin\theta}{bcos\theta}=\frac{-b}{aesin\theta}
$

and grad $PS=\frac{bsin\theta}{acos\theta + ae}$

now im assuming i use the grad QS=grad normal to show the denominator of grad PS=0 but i cant seem to get it....
• March 24th 2010, 11:23 PM
Fyor
*sorry didn't mean to post can i delete this post?
• March 24th 2010, 11:47 PM
Fyor
"Evaluate the following integral:"
zz...same mistake
• March 25th 2010, 12:22 AM
earboth
Quote:

Originally Posted by jiboom
it's late and i cant get these 2 question to fall out! help please :)

I have point $p(\bold{\color{blue}3\sqrt2 cos \theta},3sin \theta)$ on ellipse

$
\frac{x^2}{18}+\frac{y^2}{9}=1
$

I have the tangent and normal to this ellipse meet Y-axis at T and N. Then i found the centre of the circle through T,N,P to be

$
X=(0,\frac{3}{2}[\frac{1-sin^2\theta}{sin\theta}]
$

The last part asks for the equation of the locus of Q on PX produced such that X is the midpoint of PQ

so ive let Q=(x,y) then

$
\frac{x+3\sqrt(2)}{2}=0
$

$
\frac{y+3sin\theta}{2}=\frac{3}{2}[\frac{1-sin^2\theta}{sin\theta}]
$

so $Q=(\bold{\color{blue}-3\sqrt2 cos \theta},3[\frac{1-2sin^2\theta}{sin\theta}])$

but i cant find the locus. The book says

$
\frac{2x^2}{9}+\frac{9}{16y^2}=1
$

....

Since PQ is a diameter of the circle around X whose x-coordinate is allways zero the x-coordinate of Q must contain the cos-function of $\theta$.

The parametric equation of the locus is subsequently:

$L:\left\{\begin{array}{l}x= -3\sqrt2 cos \theta \\ y= 3\left(\frac{1-2sin^2\theta}{sin\theta}\right)\end{array}\right.$

Now eliminate the parameter
• March 25th 2010, 04:11 AM
jiboom
Quote:

Originally Posted by earboth
Since PQ is a diameter of the circle around X whose x-coordinate is allways zero the x-coordinate of Q must contain the cos-function of $\theta$.

The parametric equation of the locus is subsequently:

$L:\left\{\begin{array}{l}x= -3\sqrt2 cos \theta \\ y= 3\left(\frac{1-2sin^2\theta}{sin\theta}\right)\end{array}\right.$

Now eliminate the parameter

Thanks for the reply. I just realised my typo and came back to edit it. What you have posted is what i have got but i cant seem to get

$
\frac{2x^2}{9}+\frac{9}{16y^2}=1

$