If I have a conic

$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$

and want to know the tangent line at $(x_0, y_0)$, I thought I would just find the derivative (implicit) and use the equation of a line $(y-y_0) = m (x-x_0)$. For the derivative I get:

$$2Ax + By + \frac{dy}{dx}Bx + 2Cy\frac{dy}{dx} + D + E \frac{dy}{dx} =0$$

Solving for $\frac{dy}{dx}$ I get

$$m = \frac{dy}{dx} = \frac{-2Ax - By - D}{Bx + 2Cy + E}$$

Substituting this into the equation for a line:

$$(y-y_0) = \frac{-2Ax - By - D}{Bx + 2Cy + E} (x-x_0) = $$

and grouping the $x$, $y$, and constant terms, I get

$$(2Ax_o+By_0+D)x + (Bx_0+2Cy_0+E)y -2Bx_0y_0 - 2Cy_0^2 - Ey_0 - 2Ax_0^2 - Dx_0 = 0 $$

I've seen things like this: http://demonstrations.wolfram.com/TangentLinesToAConicSection/ that show an $F$ in the expression. The $x$ and $y$ terms are the same, but the constant term is $-Ax_0^2 - Bx_0y_0 - Cy_0^2 + F$. In my method here, the $F$ term goes away in the differentiation immediately as it is a constant. Can anyone explain how these representations are equivalent?

$$-Ax_0^2 - Bx_0y_0 - Cy_0^2 + F = - 2Cy_0^2 - Ey_0 - 2Ax_0^2 - Dx_0$$

up vote -4 down vote accepted

You have missed $-2Bx_0y_0$ at the Right hand side.

As $(x_0,y_0)$ lies on the curve, it will satisfy the equation of the given curve

Hence, $$Ax_0^2 + Bx_0y_0 + Cy_0^2 + Dx_0 + Ey_0+ F = 0$$

$$-Ax_0^2 - Bx_0y_0 - Cy_0^2 + F = - 2Cy_0^2 -2Bx_0y_0 - Ey_0 - 2Ax_0^2 - Dx_0$$ $$\iff Ax_0^2 + Bx_0y_0 + Cy_0^2 + Dx_0 + Ey_0+ F = 0$$

  • Sorry, I don't follow - sure, $(x_0, y_0)$ lies on the curve, but how does that make $-Ax_0^2 - Bx_0y_0 - Cy_0^2 + F$ equal to $ - 2Cy_0^2 - Ey_0 - 2Ax_0^2 - Dx_0$ ? – David Doria Mar 28 '13 at 15:58
  • @DavidDoria, make the right hand side $=0$ by taking all the terms in the Left Hand side – lab bhattacharjee Mar 28 '13 at 16:00
  • Ah, I see, I see - you simply rearranged the terms after substituting $(x_0, y_0)$. Thank you - this was driving me crazy. – David Doria Mar 28 '13 at 16:04
  • @DavidDoria, my pleasure. Hope I could make the point clear. – lab bhattacharjee Mar 28 '13 at 16:06
  • 1
    @labbhattacharjee: $$Ax_0^2 + Bx_0y_0 + Cy_0^2 + Dx_0 + Ey_0+ F = 0$$ is not the equation of tangent at point $(x_0,y_0)$. Your equation just has constant terms. Where are the variables? $(x_0,y_0)$ are constants, and $A,B,C,D,E,F$ are also constants. – Manoj Pandey Mar 28 '13 at 16:27

Theory: Equation of tangent at any point of the conic is given by $T = 0$.

$T$ is the transformed equation of the conic.

Following transformations are considered while transforming a conic at a point $(x_0, y_0)$

$$x^2 \implies x x_0$$

$$y^2 \implies y y_0$$

$$x \implies \frac{x+x_0}{2}$$

$$y \implies \frac{y+y_0}{2}$$

$$xy \implies x.y_0+y.x_0$$


Answer to your question: According to your question, the equation of conic is:

$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$

Tangent at the point: $(x_0,y_0)$ will be given by $T$= $0$.

$\implies$ $T$=$0$

$\implies$ $$A(x.x_0) + B(x.y_0+y.x_0) + C(y.y_0) + D.\frac{(x+x_0)}{2} + E.\frac{(y+y_0)}{2} + F = 0$$

This is required equation of tangent.

  • have you discovered the Question you are pointing to? – lab bhattacharjee Mar 28 '13 at 16:37

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