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Quisquis
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Homework Statement
Let [tex]\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1[/tex] be the equation of an ellipse with vertices ([tex]\pm[/tex]a,0) and (0,[tex]\pm[/tex]b), and a slope m= -0.5 in the point (1,1).
Find a and b.
The Attempt at a Solution
So far I've solved for a and b in relation to each other, but I'm not sure how to solve them specifically.
[tex]\frac{d}{dx}\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\right)[/tex]
which led me to:
[tex]\frac{2x}{a^{2}}+\frac{2yy'}{b^{2}}=0[/tex]
solving for a and b:
[tex]b=\sqrt{-\frac{2yy'a^{2}}{2x}}[/tex][tex]a=\sqrt{-\frac{2xb^{2}}{2yy'}}[/tex]
simplifying using the point (1,1) for x and y and -0.5 for y' I got:
[tex]b=\frac{a}{\sqrt{2}}[/tex]
[tex]a=b\sqrt{2}[/tex]
Substituting [tex]b\sqrt{2}=a[/tex] for [tex]a^{2}[/tex] in the derivative eq, I got:
[tex]\frac{2x}{2b^{2}}+\frac{2yy'}{b^{2}}=0[/tex]
plugging in the point (1,1) for x and y and y'=-0.5
[tex]\frac{1}{b^{2}}-\frac{1}{b^{2}}=0[/tex]
I get the same kind of thing for a, which is nice but it doesn't seem useful.
Using the same substitution in the original eq, I got [tex]b=\sqrt{\frac{3}{2}}[/tex]. I'm not sure if that's true, but if it is then I know I can do that in the same way for a.
Some help if I'm on the wrong track would be appreciated, and if I'm on the right track I'd love to hear that too.
Thanks very much for your help guys.
EDIT: solving for a in the original eq, I got [tex]a=\sqrt{3}[/tex]
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