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sandman1 Group Title

Let E be the ellipse given by the equation X^2+7y^2=8 1) if m is any real number, find all tangent lines to E that pass throught the point (m,0) 2) the ellipse E has a tangent line with postive slope that passes throught the point (-8,0) find the point of intersection of this line with the vertical line at x=13 need help please

  • 2 years ago
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  1. satellite73 Group Title
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    i guess we need the derivative first

    • 2 years ago
  2. satellite73 Group Title
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    \[2x+14yy'=0\] \[y'=-\frac{x}{7y}\]

    • 2 years ago
  3. 2bornot2b Group Title
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    This is a wonderful problem satellite

    • 2 years ago
  4. 2bornot2b Group Title
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    Take a look, the point (m,0) may not lie on the ellipse, are you getting the complexity?

    • 2 years ago
  5. satellite73 Group Title
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    oh then i probably messed up somewhere

    • 2 years ago
  6. 2bornot2b Group Title
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    I didn't mean you messed, I just wanted to let you know that the question is good

    • 2 years ago
  7. sandman1 Group Title
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    ive been trying to get this problem for the last like hour

    • 2 years ago
  8. 2bornot2b Group Title
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    Let \(y=mx+c\) be the tangent to the ellipse and let the point \((m,0) \) also lie on it

    • 2 years ago
  9. satellite73 Group Title
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    so we have to find the equation of the line through (m,0) that touches the ellipse right?

    • 2 years ago
  10. 2bornot2b Group Title
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    Right

    • 2 years ago
  11. satellite73 Group Title
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    \[y=cx-m\]since we cannot use m for the slope

    • 2 years ago
  12. 2bornot2b Group Title
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    Right. we can't use m, since its already up there

    • 2 years ago
  13. satellite73 Group Title
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    and we have to make sure that \[c=-\frac{x}{7y}\] as well so perhaps we will get two equations

    • 2 years ago
  14. 2bornot2b Group Title
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    OK, so lets assume that the line \(y=kx+c\) is a tangent to the given ellipse

    • 2 years ago
  15. satellite73 Group Title
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    i am lost already

    • 2 years ago
  16. 2bornot2b Group Title
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    So we know c must be equal to \(\pm \sqrt{a^2k^2+b^2}\)

    • 2 years ago
  17. satellite73 Group Title
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    oh oopos

    • 2 years ago
  18. 2bornot2b Group Title
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    Where a and b are from the ellipse

    • 2 years ago
  19. satellite73 Group Title
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    you are way ahead of me. i only know that the line must look like \[y=c(x-m)\] where c is the slope. i also know that we must have \[c=-\frac{x}{7y}\]

    • 2 years ago
  20. 2bornot2b Group Title
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    Yes, you are right. And I was thinking of considering that at the end. Because we have two conditions to consider. First the line is a tangent And the second is what you are doing, i.e. the point lies on that line

    • 2 years ago
  21. 2bornot2b Group Title
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    So what I am stating up there, is the condition for the line \(y=kx+c\) to be a tangent to that ellipse

    • 2 years ago
  22. 2bornot2b Group Title
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    And I am considering the general ellipse, \(\huge \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

    • 2 years ago
  23. satellite73 Group Title
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    ok i have this, tell me what you think

    • 2 years ago
  24. satellite73 Group Title
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    we know the line has slope \[-\frac{x}{7y}\] and it passes through (m,0) meaning it has the equation \[y=-\frac{x}{7y}(x-m)\] now we get \[7y^2=-x(x-m)\] \[7y^2=-x^2+xm\] \[x^2+7y^2=xm\] and we also know that on the ellipse \[x^2+7y^2=8\] making \[xm = 8\]

    • 2 years ago
  25. 2bornot2b Group Title
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    So the if the line y=kx+c is a tangent to \(\huge \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) then by theorem we know \[c=\pm \sqrt{a^2k^2+b^2}\]

    • 2 years ago
  26. 2bornot2b Group Title
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    So the equation of the line is \[y=kx+\pm \sqrt{a^2k^2+b^2}\]

    • 2 years ago
  27. 2bornot2b Group Title
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    Also since the line passes through the point (m,0) \[0=km\pm \sqrt{a^2k^2+b^2}\]

    • 2 years ago
  28. 2bornot2b Group Title
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    The above equation will give you two values of k, by solving the quadratic. Solve them, and put them in the actual equation to the st line, and you will get the two tangents

    • 2 years ago
  29. sandman1 Group Title
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    ok thanks

    • 2 years ago
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