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Math Help - Conics - Ellipse

  1. #1
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    Could someone please please give me hints on how to do these questions??? I've been trying to work them out all day! PLEASE!!!!!

    1.The tangent at P on the ellipse x^2/a^2 + y^2/b^2 =1 cuts the x axis at T and the perpendicular PN is drawn to the x-axis. If O is the origin, prove that ON * OT = a^2.

    2. Find the equation of the normal l to 9x^2 + 25y^2 = 225 at P(3, 2 2/5). This normal cuts the x-axis at G and N is the foot of the perpendicular drawn from P to the x-axis. Find GN.

    3. Find the equation of the tangent at (4,-1) to the ellopse 9x^2 + 25y^2 = 169. Prove that the circle x^2 + y^2 + 28x - 23y = 152 touches the ellipse at this point.

    4. FInd the equations of the normal to x^2 + 4y^2 = 100 at P(8,3). If the normal at P meets the major axis in G, and OY is the perpendicular from O to the tangent at P, prove that PG * OY is equal to the square on the minor semi-axis.</SPAN>
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    1.The tangent at P on the ellipse x^2/a^2 + y^2/b^2 =1 cuts the x axis at T and the perpendicular PN is drawn to the x-axis. If O is the origin, prove that ON * OT = a^2.
    Take P to be the point (a\cos\theta,b\sin\theta). Then ON is just the x-coordinate of P, namely a\cos\theta. The tangent at P has equation \frac{x\cos\theta}a + \frac{y\sin\theta}b = 1. Find where that crosses the x-axis, and that will tell you the length OT.
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    Senior Member bkarpuz's Avatar
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    Hi xwrathbringerx,

    Below, you will find solutions to Problems 1 and 2.

    Quote Originally Posted by xwrathbringerx View Post
    1.The tangent at P on the ellipse x^2/a^2 + y^2/b^2 =1 cuts the x axis at T and the perpendicular PN is drawn to the x-axis. If O is the origin, prove that ON * OT = a^2.
    Consider the ellipse
    \mathcal{E}:\ \bigg(\frac{x}{a}\bigg)^{2}+\bigg(\frac{y}{b}\bigg  )^{2}=1.
    Let P=(p,q)\in\mathcal{E}, then the equation of the tangent line at the point P is given by
    y=-\frac{b^{2}p}{a^{2}q}(x-p)+q
    or eqivalently
    a^{2}qy+b^{2}px=a^{2}b^{2}.......................(1)
    Actually, obtaining the equation of the tangent line is not very simple (just makes use of a simple idea but too much computations) but If you need I can also show it.
    To find T=(t,0), we have to plug T into (1), i.e., put y=0 and x=t.
    Therefore, we get
    t=\frac{b^{2}a^{2}}{b^{2}p}=\frac{a^{2}}{p}..............................(2)
    On the other hand, if N=(n,0) is the projection of P=(p,q), we must have
    n=p..........................................(3)
    And finally, from (2) and (3), we can find that
    |ON|\times|OT|=n\times t=p\times\frac{a^{2}}{p}=a^{2}
    As a note, the value \big||OT|-|ON|\big| is called the subtangent length of the point P (see Figure 1). \rule{0.2cm}{0.2cm}

    Quote Originally Posted by xwrathbringerx View Post
    2. Find the equation of the normal l to 9x^2 + 25y^2 = 225 at P(3, 2 2/5). This normal cuts the x-axis at G and N is the foot of the perpendicular drawn from P to the x-axis. Find GN.
    You can see from the solution of Problem 1 that the slope of the tangent line at P\in\mathcal{E} is
    m_{T}=-\frac{b^{2}p}{a^{2}q}.
    Hence, and the point P the normal line must have the slope
    m_{N}=\frac{a^{2}q}{b^{2}p}
    since m_{T}m_{N}=-1.
    Then, the equation of the normal line is at the point P\in\mathcal{E} is
    y=\frac{a^{2}q}{b^{2}p}(x-p)+q
    or equivalently
    a^{2}qx-b^{2}py=c^{2}pq,......................(4)
    where
    c^{2}:=a^{2}-b^{2}.
    To find G=(g,0), use (4) with x=g and y=0, then we get
    g=\frac{c^{2}pq}{a^{2}q}=\frac{c^{2}p}{a^{2}}............................(5)
    And the foot N=(n,0) must satisfy
    n=p..........................................(6)
    So that (5) and (6) yield
    |GN|=\big||OG|-|ON|\big|=\bigg|\frac{c^{2}p}{a^{2}}-p\bigg|=\bigg|\bigg(\frac{c^{2}}{a^{2}}-1\bigg)p\bigg|
    ......_ =\bigg|\frac{c^{2}-a^{2}}{a^{2}}p\bigg|
    ......_ =\frac{b^{2}}{a^{2}}|p|.
    This value is called the subnormal length of the point P (see Figure 1).
    You may put above a=5, b=3 and p=3 to get your answer. \rule{0.2cm}{0.2cm}


    Figure 1: Subtangent, and subnormal lengths.

    Think of the remaining ones with drawing figures,
    if you wont be able to succeed I can give a hand again.
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    Hi Opalg and bkarpuz

    Thanks guys for helping me out. I can't believe i can't figure out these last few questions off the worksheet.

    I've tried looking at those last 2 questions but ... . Sorry to be such a bother but could you please guide me in those too?
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    Quote Originally Posted by xwrathbringerx View Post
    I can't believe i can't figure out these last few questions off the worksheet.

    I've tried looking at those last 2 questions but ... . Sorry to be such a bother but could you please guide me in those too?
    Some general comments about ellipses: If the point P(x_0,y_0) lies on the ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 then the equation of the tangent at P is \frac{x_0x}{a^2} + \frac{y_0y}{b^2} = 1. That's an easy equation to remember — you just take the equation of the ellipse and change one of the x's to x_0 and one of the y's to y_0. For many problems, it's useful to use the parametric form P = (a\cos\theta,b\sin\theta) for a point on the ellipse. The above equation for the tangent then becomes \frac{x\cos\theta}a+\frac{y\sin\theta}b = 1.

    The slope of the tangent is thus -\frac{b^2x_0}{a^2y_0}. So the slope of the normal is \frac{a^2y_0}{b^2x_0}, and the equation of the normal is y-y_0 = \tfrac{a^2y_0}{b^2x_0}(x-x_0). (That equation is probably not worth remembering, but you should remember how to obtain it.)

    Quote Originally Posted by xwrathbringerx View Post
    3. Find the equation of the tangent at (4,-1) to the ellipse 9x^2 + 25y^2 = 169. Prove that the circle x^2 + y^2 + 28x - 23y = 152 touches the ellipse at this point.
    You can write the equation of the tangent at P(4,–1) (remembering the formula above) as 36x - 25y = 169, and the normal at P is 25x+36y=64. The equation of the circle can be written as (x+14)^2 + (y-23/2)^2 = \text{(some large constant)}, so its centre is at (-14,23/2). Check that P lies on the circle, and that the centre of the the circle lies on the normal at P. Then think about how those two facts imply that the circle and the ellipse touch at P.

    Quote Originally Posted by xwrathbringerx View Post
    4. FInd the equations of the normal to x^2 + 4y^2 = 100 at P(8,3). If the normal at P meets the major axis in G, and OY is the perpendicular from O to the tangent at P, prove that PG * OY is equal to the square on the minor semi-axis.</SPAN>
    Tangent at P is 8x + 12y = 100, with slope –2/3. Normal has slope 3/2, and equation y-3 = \tfrac32(x-8), or 2y-3x+18=0. This meets the major axis when y = 0, so that x = 6, and G is the point (6,0). The perpendicular from O to the tangent at P will be perpendicular to the tangent and therefore parallel to the normal; and it will pass through the origin. So its equation is 2y – 3x = 0.

    From that, you can work out the location of Y, and the distances PG and OY. Finally, to get the minor semi-axis b, write the equation of the ellipse as \frac{x^2}{100} + \frac{y^2}{25} = 1, from which you can read off a^2 and b^2.
    Last edited by Opalg; February 21st 2010 at 08:51 AM. Reason: typo
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    Check that P lies on the circle, and that the centre of the the circle lies on the normal at P. Then think about how those two facts imply that the circle and the ellipse touch at P.
    Opalg. I don't seem to get what the proofs imply ...

    Is it because it demonstrates that they share a common tangent?
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  7. #7
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    Quote Originally Posted by xwrathbringerx View Post
    I don't seem to get what the proofs imply ...

    Is it because it demonstrates that they share a common tangent?
    That's right. The circle and the ellipse have the same normal, and therefore the same tangent, at the point (4,1), and that is what is meant by saying that they touch at that point.
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