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sandman1

  • 2 years ago

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

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

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

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

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

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

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

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

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

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

  10. 2bornot2b
    • 2 years ago
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    Right

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

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

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

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

  15. satellite73
    • 2 years ago
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    i am lost already

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

  17. satellite73
    • 2 years ago
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    oh oopos

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

  19. satellite73
    • 2 years ago
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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 \[c=-\frac{x}{7y}\]

  20. 2bornot2b
    • 2 years ago
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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

  21. 2bornot2b
    • 2 years ago
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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

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

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

  24. satellite73
    • 2 years ago
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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 \[xm = 8\]

  25. 2bornot2b
    • 2 years ago
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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 \[c=\pm \sqrt{a^2k^2+b^2}\]

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

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

  28. 2bornot2b
    • 2 years ago
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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

  29. sandman1
    • 2 years ago
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    ok thanks

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