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Y axis intercepts of ellipse tangents

  1. Mar 5, 2016 #1
    1. The problem statement, all variables and given/known data
    It is given that the line y= mx + c is a tangent to the ellipse

    \frac{x^2}{a^2} + \frac{y^2}{b^2}=1[/itex] if [itex]a^2m^2=c^2-b^2[/itex]

    Show that if the line y=mx+c passes through the point (5/4, 5) and is tangent to the ellipse [itex]8x^2+3y^2=35[/itex], then c = 35/3 or 35/9

    2. Relevant equations

    3. The attempt at a solution
    The tangents should be in the form
    [itex]y=\pm \frac{c^2-b^2}{a^2}x+c[/itex]
    I tried substituting this value for y into the equation of the given ellipse but it became a bit messy so I deployed a different tact:
    The intersection of the tangents is equal to (5/4, 5), so multiplying the tangent equations together should give the point of intersection enabling me to solve for c

    c=\frac{(x(\pm \sqrt{a^2y^2-a^2b^2+b^2x^2}))+ay^2}{a^2-x^2}

    Substituting a=35/3 and b=35/8 gives 7.719285513 and 3.827106245.

    When I insert these values into a graph drawing app they seem approximately correct whereas one of the given values in the question, 35/3, does not as far I can tell.

    Are the values for c given in the question correct. Are my values for c correct?
  2. jcsd
  3. Mar 5, 2016 #2


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    How do you multiply tangent equations together and what do you expect as result?

    You know a and b, so you get an equation involving m and c and known constants only.
    The point (5/4, 5) gives another equation with m and c only. This one is linear, so I would use this and plug it into the other one. Ideally, you get the right two solutions for c.

    7.719285513 and 3.827106245 are not 35/3 and 35/9.
  4. Mar 6, 2016 #3
    Hi mfb, thanks for your reply. Do you mean that I should plug [itex]m=4-\frac{4}{5}c[/itex] into [itex]\frac{35}{3}m^2=c^2-\frac{35}{8}[/itex]?

    When I do this I get values for c identical to those I had previously (although perhaps a bit more efficiently this time):

    c=7.719285508 or 3.827106245
    This suggests to me that multiplying the tangent equations was legitimate if a little unnecessary. The aim was to eliminate the square root, which your method bypasses quite successfully.

    My main question now is should I bin my book. Are the values for c stated in the question correct?

    Sorry if I've misinterpreted your guidance, I'm not ruling out this possibility.
  5. Mar 6, 2016 #4


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    I think you swapped the values for a and b.
  6. Mar 6, 2016 #5
    Isn't the convention to assign [itex] a[/itex] to the value of the semi major axis? In this case 35/3 > 35/8 so shouldn't a=35/3 and b=35/8?

    To check I'm not mistaken I repeated the procedure above, swapping a and b, and ended up with the square root of a negative number.
  7. Mar 6, 2016 #6


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    The order of a and b does not matter, but it has to be consistent within the problem. The problem divides x^2 by a^2, so you cannot call that b^2.

    WolframAlpha finds nice solutions - exactly those given.
  8. Mar 6, 2016 #7
    Looks like the negative square root i produced above was an error. Having gone through it again I am able to produce the values for c in the question. Thank you for identifying my error in mixing up a and b.
  9. Mar 6, 2016 #8
    I am just curious. For which class is this?, Real analysis?
  10. Mar 6, 2016 #9
    The chapter of the book is called coordinate geometry. I'm not sure if there is a more appropriate title for this branch of maths.
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