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Ellipse perpendicular distance

  1. Feb 15, 2009 #1
    1. The problem statement, all variables and given/known data


    Prove that the straight line x cos @ + y sin @ = p is a tangent to the ellipse x2/a2 + y2/b2 if a2 cos2@ + b2 sin2@ =p2 .
    u and v are the perpendicular distances of a tangent from the two points M(0,ae) and N(0,-ae) respectively. Prove that u2 + v2 is a constant.


    2. The attempt at a solution

    So far I have: xx1/a2+yy1/b2 -1 = x cos@ + y sin@ - p
    I don't know where to go from here and I know in the second part I have to use the perpendicular distance formula but I'm not sure what to sub in and why.
     
    Last edited: Feb 15, 2009
  2. jcsd
  3. Feb 15, 2009 #2

    HallsofIvy

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    You mean the elliplse [tex]x^2/a^2+ y^2/b^2=1[/tex]. You can simplify that a little by writing it as [tex]b^2x^2+ a^2y^2= a^2b^2[/tex]. Differentiating with respect to x, 2b^2x+ 2a^2yy'= 0 so y'= -b^2x/a^2y. The tangent line at the point [tex](x_1, y_1)[/tex] is given by [tex]y= -(b^2x_1/a^2y_1)(x- x_1)+ y_1[/tex] or b^2x_1x+ a^2y_1y= b^2x_1^2+ a^2y_1^2[/tex].

    If we are going to be able to write the coefficients as [tex]cos(\theta)[/itex] and [itex]sin(\theta)[/itex], we must We must have [tex]cos^2(\theta)+ sin^2(\theta)= 1[/tex]. Now we have [tex]b^4x_1^2+ a^4y_1^2[/tex] as the sum of the squares of the coefficients so divide the entire equation by the square root of that:
    [tex]\frac{b^2x_1}{\sqrt{b^4x_1^2+ a^4y_1^2}}x+ \frac{a^2y_1}{\sqrt{b^4x_1^2+ a^4y_1^2}}y= \frac{b^2x_1^2+ a^2y_1^2}{\sqrt{b^4x_1^2+ a^4y_1^2}}[/tex]

    That is of the form [tex]cos(\theta)x+ sin(\theta)b= p[/tex]
     
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