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Find the coordinates of those points on the curve given by the equation:

$\ x^2-0.25xy+y^2=16 $

at which the tangent line has slope 1.

Implicitly derived:$\ \frac{(-2x+0.25y)}{(2y-0.25x)} $

The book talked something about solving for y (from the original equation):

$\ \pm \sqrt{-x^2+0.25xy+16}$

then substituting it back into the implicitly derived equation and solving, but I am not sure that is going to work or else it is going to be very cumbersome. Anyone know how to solve this, perhaps in a simpler (possibly more correct) fashion? Thanks!

P.S. I spent 30mins figuring out the mathTex stuff so it looks pretty, it was the least I could do I suppose? Only for all you picky math geniuses who are going to help me! :)

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Try polar coordinates $x = r cos(t)$ and $y = r sin(t)$. Solve for $r$ and you'll obtain a function of $t$. This function I think will not be difficult to differentiate. I'd didn't check it, so I write a comment and not an answer... – shamovic Mar 9 '11 at 21:58
It would be good to make your implicit derivative an equation by putting $\frac{dy}{dx}$ on the other side of an equals sign. But good on you for trying, showing what you have done, and working on $\LaTeX$ – Ross Millikan Mar 9 '11 at 22:40
up vote 2 down vote accepted

Since it's an equation of degree 2, the curve in question is a conic section. Looking at the coefficients, it's probably an ellipse. (The criterion for that is that the quadratic form should always be positive.)

The equation is symmetric in $x$ and $y$. In other words, a reflection at the line $y=x$ will map the ellipse to itself. This also means that the line $y=x$ is also a main axis.

Ellipse has main axis $y=x$

Coincidentally, the question also asks for a tangent with exactly the same slope as the main axis. This can only happen at the points where the other main axis intersects the ellipse. In our case, the other axis is the line $y=-x$.

Hence, the points in question have the form $(x,-x)$ and can be obtained by solving the equation

$$ x^2 + 1/4·x^2 + x^2 = 16 $$

i.e. $x = \pm\sqrt{64/9} = \pm 8/3$.

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Thanks, you even drew me a nice picture! Makes so much more sense now! Stupid calculus books make everything so complicated... – Mr_CryptoPrime Mar 9 '11 at 22:33
Of course, I didn't actually use calculus to solve the problem. :-) If the tangent slope would have been anything other than 1, then my argument wouldn't work, but the implicit differentiation thing would still work. – Greg Graviton Mar 10 '11 at 9:20

You have already implicitly derived the expression for $\frac{dy}{dx}$. You can set that equal to 1 and simplify to get $x = -y$. Substitute that in the original equation and you have a quadratic that can be easily solved.

I got the solutions $x = \pm 8/3$ and $y = \mp 8/3$.

The book seems to be making a big deal out of a very simple question.

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