Why does the "T=0" method to calculate tangent work?

Question

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?

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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))

Answer 1

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.

Answer 2

First of all, you probably know that the tangent at $x_0$ of the parabola $y=ax^2$ is: $$ y = ax_0^2 + 2ax_0(x-x_0). $$

If you didn't know this, keep reading! This means that, if we stay "close enough" to $x_0$, the line above is the one that approximates best the direction of the parabola.

In particular, if we move from $x_0$ to $x_0 + \Delta x$, we get: $$ a(x_0+\Delta x)^2 = ax_0^2 + 2ax_0\Delta x + a(\Delta x)^2. $$

You see that if $\Delta x$ is very small, the last term tends to vanish, and $$ a(x_0+\Delta x)^2 \approx ax_0^2 + 2ax_0\Delta x $$

becomes a very good approximation. Since $\Delta x = x-x_0$, the equation of the line given by: $$ y = ax_0^2 + 2ax_0(x-x_0) $$

is the equation of our desired tangent.

Your "T-formula" implies that a curve $C$ with equation: $$ ax^2 + 2hxy + by^2 + 2 gx +2 fy + c =0 $$

has tangent $T$ in $(x_0,y_0)$: $$ axx_0 + h(xy_0+x_0y) + byy_0 + g(x+x_0) + f(y+y_0) +c=0. $$

We would like to show now that $T$ is tangent, that is, that $T$ is the best possible approximation of $C$ if we stay "close enough" to the point.

If we move from $(x_0,y_0)$ to $(x_0+\Delta x, y_0+\Delta y)$, like before, we get: $$ a(x_0+\Delta x)^2 + 2h(x_0+\Delta x)(y_0+\Delta y) + b(y_0+\Delta y)^2 + 2 g(x_0+\Delta x) +2 f(y_0+\Delta y) + c =0. $$

Expanding the products: $$ \begin{array}{c} a(x_0^2+2x_0\Delta x+\Delta x^2)+ \\ + 2h(x_0y_0+\Delta xy_0+x_0\Delta y+\Delta x\Delta y)+ \\ + b(y_0^2+2y_0\Delta y+\Delta y^2)+ \\ + 2 g(x_0+\Delta x) +\\ +2 f(y_0+\Delta y) + c =0. \end{array} $$

Like before, we can neglect the quadratic "small" terms to get a linear equation, which approximates our curve in the best way possible. We obtain: $$ \begin{array}{c} a(x_0^2+2x_0\Delta x) + 2h(x_0y_0+\Delta xy_0+x_0\Delta y) + b(y_0^2+2y_0\Delta y)+ \\ + 2 g(x_0+\Delta x) +2 f(y_0+\Delta y) + c =0. \end{array} $$

To get the equation in $x,y$ for the line, we must (like before) replace $\Delta x$ with $(x-x_0)$, and now also $\Delta y$ with $(y-y_0)$. So: $$ \begin{array}{c} a(x_0^2+2x_0(x-x_0)) + 2h(x_0y_0+(x-x_0)y_0+x_0(y-y_0)) + b(y_0^2+2y_0(y-y_0))+ \\ + 2 g(x_0+x-x_0) +2 f(y_0+y-y_0) + c =0. \end{array} $$

Summing, we get: $$ \begin{array}{c} a(-x_0^2+2x_0x) + 2h(-x_0y_0+xy_0+x_0y) + b(-y_0^2+2y_0y)+ \\ + 2 gx +2 fy + c =0. \end{array} $$

This is an equation for the line, but we still haven't used the fact that the line must pass through our point $(x_0,y_0)$. Or, we still have to ensure that $(x_0,y_0)$ satisfies the equation for $C$. Now, we rewrite the expression above in the following way (we have moved the terms in $a,h,c$, and added and subtracted the terms in g,f,c): $$ \begin{array}{c} a(2x_0x) + 2h(xy_0+x_0y) + b(2y_0y) + g(x +x_0) + f(y+y_0) + 2c +\\ -(ax_0^2+2hx_0y_0+by_0^2+gx_0+fy_0+c) =0. \end{array} $$

Saying that $(x_0,y_0)$ passes through $C$ is saying that the second line vanishes! So we are finally left with: $$ 2ax_0x + 2h(xy_0+x_0y) + 2by_0y + g(x +x_0) + f(y+y_0) + 2c =0. $$

Dividing both sides by $2$, we get exactly your formula.

(Yes, it is a rather long and complicated process. I can't come up with a simpler one that doesn't lose clarity and doesn't use advanced calculus. This is probably why they never prove it in high school! Anyway, I'd be very very happy to see a shorter, elementary proof.)

Answer 3

Replying to OP's answer: (let us keep the original 2 degree curve equation to be S and when you substitute ($x_1,y_1$), let it be be $S_1$). Also, I am pretty sure OP might have understood his mistake till now, but just for clarification, his solution is incorrect. In his solution, he has taken $\frac{dy}{dx}$ to be in terms of x,y and substituted them in the equation $(y-y_1) = \frac{dy}{dx}(x-x_1)$, while you have to substitute them in terms of $(x_1,y_1)$, what I mean is that they should not be kept in general terms (x,y) (line equation contains y = mx+c where (x,y) are general points on line) and should be dependent upon point of contact of tangent to the curve. He then proceeded to open the equation $(y-y_1) = \frac{dy}{dx}(x-x_1)$ where $\frac{dy}{dx}$ was still in (x,y) terms and then added gx + fy + c on both sides and used $S_1 = 0$, which is wrong because gx + fy + c is a general point in 2D space and not necessarily on the curve (because he confused $(x_1,y_1)$ with (x,y) in the original portion of the equation $((y-y_1) = \frac{dy}{dx}(x-x_1))$ he could successfully do this and get the final answer as $T = 0)$; The real solution is along the same lines:

$\frac{dy}{dx} = -\frac{(ax_1 +hy_1 + g)}{(hx_1 + by_1 + f)}$

Then using $(y-y_1) = \frac{dy}{dx}(x-x_1)$ and opening up the equation, we get: $hyx_1 + byy_1 + fy - hx_1y_1 - b(y_1)^2 - fy_1 = -axx_1-hxy_1-gx+a(x_1)^2 + hx_1y_1 + gx_1$

Rearrange to get $axx_1 + byy_1 + hxy_1 + hyx_1 + gx + fy = a(x_1)^2 + b(y_1)^2 + 2hx_1y_1 + gx_1 + fy_1$

Now you can add $gx_1 + fy_1 + c$ (on both sides as they are constants) to get (by using $S_1$ = 0) $T = 0$;

As for the comment below OP's answer by @Sufaid Saleel as to what would happen if $hx_1 + by_1 + f = 0$ (then we obviously cannot divide by it to get $\frac{dy}{dx} = -\frac{(ax_1 +hy_1 + g)}{(hx_1 + by_1 + f)}$

So what we could do is take 2 cases, one where $ax_1 + hy_1 + g = 0$ and one where it is $\neq 0$ :

For the first case, a little bit of manipulation(multiplying first equation by $x_1$ and second equation by $y_1$ and adding and using $a(x_1)^2 + b(y_1)^ + 2hx_1y_1 .. = 0$ (this allows for more cases like $x_1 = 0 \& y_1 = 0$ and stuff like that, but they are cases which can be easily solved) shows that Δ of the given equation is $0$, implying that this is a pair of straight lines and you successfully found out the point of intersection of the two line $(x_1,y_1)$ (as you set both $\frac{∂S}{∂x}$ and $\frac{∂S}{∂y} = 0$ which gives the centre of the conic (or if it is a pair of straight lines - the point of intersection of both lines))

If $ax_1 + hy_1 + g \neq 0 \implies$ there is a vertical tangent to the curve at that point (as the equation $ax_1 + hy_1 + g + \frac{dy}{dx}$$(by_1 + hx_1 + f)$ has to be equal to $0$ at that point, but the term with $\frac{dy}{dx} = 0 \implies \frac{dy}{dx}$ has to be ∞ (positive or negative) (like the curve $2x^2 + 2y^2 + 6xy + 3x + 2y + 3 = 0$ has a vertical tangent at (-1,1)).