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Math Help - co-ordinate geometry

  1. #1
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    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

     <br />
\frac{x^2}{18}+\frac{y^2}{9}=1<br />

    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

     <br />
X=(0,\frac{3}{2}[\frac{1-sin^2\theta}{sin\theta}]<br />

    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

     <br />
\frac{x+3\sqrt(2)}{2}=0<br />
     <br />
\frac{y+3sin\theta}{2}=\frac{3}{2}[\frac{1-sin^2\theta}{sin\theta}]<br />

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

    but i cant find the locus. The book says

     <br />
\frac{2x^2}{9}+\frac{9}{16y^2}=1<br />

    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}

    but QS parallel to normal so grad normal= grad QS

    so

     <br />
\frac{asin\theta}{bcos\theta}=\frac{-b}{aesin\theta}<br />

    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....
    Last edited by jiboom; March 24th 2010 at 06:21 PM.
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  2. #2
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    *sorry didn't mean to post can i delete this post?
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  3. #3
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    "Evaluate the following integral:"

    zz...same mistake
    Last edited by Fyor; March 24th 2010 at 11:55 PM. Reason: reckless posting! D:
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  4. #4
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    Quote Originally Posted by jiboom View Post
    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

     <br />
\frac{x^2}{18}+\frac{y^2}{9}=1<br />

    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

     <br />
X=(0,\frac{3}{2}[\frac{1-sin^2\theta}{sin\theta}]<br />

    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

     <br />
\frac{x+3\sqrt(2)}{2}=0<br />
     <br />
\frac{y+3sin\theta}{2}=\frac{3}{2}[\frac{1-sin^2\theta}{sin\theta}]<br />

    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

     <br />
\frac{2x^2}{9}+\frac{9}{16y^2}=1<br />

    ....
    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
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  5. #5
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    Quote Originally Posted by earboth View Post
    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

     <br />
\frac{2x^2}{9}+\frac{9}{16y^2}=1<br /> <br />
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