Given a random equation of a curve: $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
Suppose we need to find the tangent to this curve at any point $A(x_1, y_1)$. A method given to me by my professor was the 'T' method:
The 'T' form of an equation can be obtained by replacing:
$$x^2 \rightarrow xx_1$$
$$y^2 \rightarrow yy_1$$
$$x \rightarrow \frac{x + x_1}{2}$$
$$y \rightarrow \frac{y + y_1}{2}$$
$$xy \rightarrow \frac{xy_1 + x_1y}{2}$$
The tangent to the curve is then the equation $T=0$.
For instance, if we need to find the tangent at $(2, 2)$ to the parabola $y^2 – 2x=0$:
$T =0$: $\implies yy_1 – 2\frac{x+x_1}{2} = 0$
Substituting $x_1 = 2$ and $y_1 = 2$: $$2y – (x+2) = 0$$ $$\implies 2y – x = 2$$
which is the required tangent.
I don't understand how this works! Could someone help me understand why it does?
This also works for other cases like:
- Deriving the equation of the two tangents to a curve from a certain external point: $SS_1 = T^2$ (S is the equation of the curve and S1 is the value given by the equation when the point is substituted into it (the power of the point wrt the curve))
Best Answer
While geodude's method works, I'm putting up my calculus version of the problem.
Taking the random curve: $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$
Differentiating it wrt x: $$2ax + 2hx\frac{dy}{dx} + 2hy + 2by\frac{dy}{dx} + 2g + 2f\frac{dy}{dx} = 0$$ $$\implies (ax + hy + g) + \frac{dy}{dx}(hx + by + f) = 0$$ $$\implies \frac{dy}{dx} = -\frac{(ax + hy + g)}{(hx + by + f)}$$
The equation of the tangent at the point $(x_1, y_1)$ is: $(y-y_1) = \frac{dy}{dx} (x - x_1)$
$$\implies (y-y_1)(hx+by+f) = (x_1 - x)(ax + hy + g)$$ $$\implies hxy + by^2 + fy -hxy_1 - byy_1 - fy_1 = axx_1 + hyx_1 + gx_1 - ax^2 - hxy - gx$$
Rearranging: $$ax^2 + by^2 + 2hxy + gx + fy = axx_1 + hyx_1 + gx_1 + hxy_1 + byy_1 + fy_1$$
Adding $gx + fy + c$ on both sides: $$ax^2 + by^2 + 2hxy + 2gx + 2fy +c = axx_1 + hyx_1 + gx_1 + hxy_1 + byy_1 + fy_1 +gx + fy + c$$
The LHS is $0$.
$$\therefore axx_1 + 2h\bigg(\frac{x_1y + xy_1}{2}\bigg) + byy_1 + 2g\bigg(\frac{x_1 + x}{2}\bigg) + 2f\bigg(\frac{y + y_1}{2}\bigg) + c = 0$$
Which is the "T" form of the equation.