# The Instant Tangent

It is interesting to note that the tangent at point $$(p,q)$$ for the circle $$(x-h)^2+(y-k)^2=r^2$$ is $$(x-h)(p-h)+(y-k)(q-k)=r^2$$ which is formulated simply by replacing one component of the squared $$(x-a), (y-b)$$ term with $$p,q$$ instead of $$x,y$$, i.e. "fixing" one of the components at $$(p,q)$$ and "releasing" the other.

Hence the tangent can be worked out instantly without going through the laborious process of finding the parametric point, differentiating to find the slope, and constructing the tangent equation!

Interestingly this seems to work for all conics (although not for other curves in general).

$$\begin{array} &&&\\ \hline \textbf{Conic}&\textbf{Equation}&\textbf{Tangent at }(p,q)\\ &\hline\\ \text{Circle} &\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{a^2}=1\qquad &\dfrac{(x-h)(p-h)}{a^2}+\dfrac{(y-k)(q-k)}{a^2}=1\qquad\\\\ \text{Ellipse}\qquad &\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1 &\dfrac{(x-h)(p-h)}{a^2}+\dfrac{(y-k)(q-k)}{b^2}=1\\\\ \text{Hyperbola}\qquad &\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1 &\dfrac{(x-h)(p-h)}{a^2}-\dfrac{(y-k)(q-k)}{b^2}=1\\\\ \text{Parabola} &\dfrac{(x-h)^2}{a^2}+\dfrac{y-k}b=1 &\dfrac{(x-h)(p-h)}{a^2}+\dfrac{\frac12 (\overline{y-k}+\overline{q-k})}b=1\\ \hline \end{array}\\$$

See desmos implementation here.

Why does this work, and is there an intuitive or geometric explanation?

Following some very useful solutions posted, here's a desmos implementation for the "Instant Tangent" of a general conic.

• As @amd reminded me in a comment to this recent answer where a version of this phenomenon occurred: If you write the conic equation as $\mathbf{x}^TC\mathbf{x}=0$, then the equation of the tangent at $\mathbf{p}$ is $\mathbf{x}^TC\mathbf{p}=0$. – Blue Jul 17 '17 at 5:14
• If you are interested, there are many more (hundreds of) shortcuts related to conics which IIT JEE aspirants study in India, including the one you mentioned and many others. – Kartik Jul 17 '17 at 9:42

There is a much simpler way to apply calculus to find these.

Let $C$ be a curve, $P$ be a nonsingular point on the curve, and $T$ be the tangent line.

If $C$ is given by an equation $u = 0$ and $T$ is given by the equation $v = 0$, then $\mathrm{d}u|_P = \mathrm{d}v|_P$.

Furthermore, if we take $v$ to be linear in $x,y$, then if we express $\mathrm{d}v$ as a linear combination of $\mathrm{d}x$ and $\mathrm{d}y$, it has constant coefficients.

Thus, knowing $\mathrm{d}u|_P$ immediately lets us determine $\mathrm{d}v$, and thus $v$ up to a constant. And it's easy to set that constant to $0$ by writing $v$ as a linear combination of terms that vanish at $P$.

Consider the example of the circle: the givens are

• $u = (x-h)^2 + (y-k)^2 - r^2$
• $P = (p,q)$

Then we compute

$$\mathrm{d}u = 2 (x-h) \mathrm{d}x + 2 (y-k) \mathrm{d} y$$ $$\mathrm{d}u|_P = 2 (p-h) \mathrm{d}x + 2 (q-k) \mathrm{d} y$$ $$v = 2 (p-h) (x-p) + 2 (q-k) (y-q)$$

and thus the tangent line is given by the equation

$$2 (p-h) (x-p) + 2 (q-k) (y-q) = 0$$

If we alternatively decided to lift $\mathrm{d}x$ to $x-h$ and $\mathrm{d}y$ to $y-k$ as you did, then the constant isn't immediately determined, although we do conclude that the tangent line is given by

$$2 (p-h) (x-h) + 2 (q-k) (y-k) = C$$

for some constant $C$. But plugging in $P$ and comparing with the equation of the circle lets us determine that $C = 2 r^2$.

• In short, the tangent hypersurface to $v=0$ at $P$ is $Dv\mid_P\cdot(x-P)=0$. – amd Jul 17 '17 at 6:44

The gradient of the circle at $(p,q)$ is $f=[2(p-h),2(q-k)]^{T}.$ This is perpendicular to the curve at the point $(p,q).$ But the tangent line is exactly the perpendicular line to $f$, so we should have $(x,y)\cdot f=(p,q)\cdot f$ for all points on the line. In other words, $2x(p-h)+2y(q-k)=2p(p-h)+2q(q-k),$ or equivalently, $(x-h)(p-h)+(y-k)(q-k)=(p-h)^{2}+(q-k)^{2}=r^{2},$ where the last equality follows since $(p,q)$ lies on the curve.

I believe this explains all of these examples.

• How would this apply in the case of the parabola? – hypergeometric Jul 17 '17 at 5:11
• $(y-k)^2-4a(x-h)=0,$ so $f=[-4a,2(q-k)]^{T},$ $(x,y)\cdot f=(p,q)\cdot f$ if and only if $(y-k)(q-k)-2a(x-h)=(q-k)^2-2a(p-h)=2a(p-h)$ since $(q-k)^2=4a(p-h)$ (since $(p,q)$ is on the curve). Moving the $x-h$ term to the other side completes this. – RideTheWavelet Jul 17 '17 at 7:14

In homogeneous coordinates, the tangent to the conic $\mathbf x^TC\mathbf x=0$ at $\mathbf p$ is $C\mathbf p$. You can find a short proof in this answer.

• You probably meant $\mathbf x^T C\mathbf p$? – hypergeometric Jul 17 '17 at 6:00
• @hypergeometric Not really. A line in $\mathbb{RP}^2$ can be represented by a homogeneous vector. So, while the Cartesian equation of the tangent at $\mathbf p$ is $(x,y,1)C\mathbf p=0$, that line can also be identified by just the vector $C\mathbf p$. – amd Jul 17 '17 at 6:43

Let $$s=A x^2 + 2Bxy + Cy^2 + 2Fx + 2Gy + H$$,

and inspired by geogebra let $$x((p,q))=p$$ and $$y((p,q))=q$$

Letting $$s_1=A x x(P_1) + B(x(P_1)y + xy(P_1)) + Cy(P_1)y + F(x(P_1) + x) + G(y(P_1) + y) + H$$ and more generally $$s_{ij} = A x(P_i)x(P_j) + B(x(P_i)y(P_j) + x(P_j)y(P_i)) + Cy(P_i)y(P_j) + F(x(P_i) + x(P_j)) + G(y(P_i) + y(P_j)) + H$$

we get your observation (proved in the link above) as:

Let point $$P_1$$ lie on the conic $$s = 0$$. In other words, assume that $$s_{11} = 0$$. Then $$s_1 = 0$$ is an equation of the line tangent to $$s = 0$$ at $$P_1$$.

Furthermore if $$P_1$$ is not assumed to be on the conic, $$s_1^2-s_{11}s=0$$ factors and defines the tangent line pair to the conic $$s=0$$ through $$P_1$$.