Partial Differentiation and Conic Asymptotes

[chaoseverlasting]

If you partially differentiate the equation of a hyperbola w.r.t. x or y do you get the equation of its asymptotes? I know that if you do partially differentiate it, the two lines that you get, intersect at its center.

This is true for any conic and pair of straight lines. What about other conics? I.e, if you partially differentiate the equations for a parabola, ellipse, circle, what exactly do you get?

 

[radou]

Investigate this link (and click 'next', too): http://www.maths.abdn.ac.uk/~igc/tch/ma1002/diff/node48.html".

 

[Hurkyl]

It might help to know that "asymptote to a curve" is essentially the same thing as "tangent line to a curve at some point at infinity".

You can shift your perspective (so that the points at infinity become ordinary points) by using a transformation, such as

(s, t) = (y/x, 1/x).

How does your hyperbola, and its asymptotes, look in (s, t)-coordinates?


Oh, if it helps, the reverse transformation is:

(x, y) = (1/t, s/t)

 

[HallsofIvy]

If you partially differentiate the equation of a hyperbola w.r.t. x or y do you get the equation of its asymptotes? I know that if you do partially differentiate it, the two lines that you get, intersect at its center.

This is true for any conic and pair of straight lines. What about other conics? I.e, if you partially differentiate the equations for a parabola, ellipse, circle, what exactly do you get?

It's not exactly clear what you mean. The equation of a hyperbola in "standard position" is
\frac{x^2}{a^2}- \frac{y^2}{b^2}= 1[/itex]<br /> If you differentiate that with respect to x and y you get, respectively,<br /> \frac{2x}{a}= 0 and<br /> \frac{2y}{b}= 0<br /> or x= 0, y= 0, certainly NOT the equations of its asymptotes, which are <br /> y= (b/a)x and y= -(b/a)x.<br /> <br /> The asymptotes of a hyperbola are the lines the hyperbola tends to for large x and y. One way of finding them is to say that, for very large x, y, &quot;1&quot; is very small and can be neglected in comparison to the other terms: the curve will satisy, approximately, for large x, y <br /> \frac{x^2}{a^2}- \frac{y^2}{b^2}= 0<br /> \frac{x^2}{a^2}= \frac{y^2}{b^2}<br /> \frac{x}{a}= \pm \frac{y}{b}[/itex]&lt;br /&gt; the equations of the asymptotes.

 

[chaoseverlasting]

Thank you, that helps a lot. I am not familiar with partial differentiation, I just know certain specific applications of it, such as the ones I mentioned above, what does it mean really?