There is a shortcut for finding the equation of a tangent to a conic. To what other curves can this shortcut be applied?

Question

The conics can be written in Cartesian and parametric form:

Conic	Cartesian equation	Parametric equation
Ellipse	$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$	$x=a \cos t, y=b \sin t$
Parabola	$y^2=4ax$	$x=at^2,y=2at$
Hyperbola	$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$	$x=a \sec t, y=b \tan t$

One task that students are frequently asked to perform is to find the equation of a tangent to a curve at a given point.

Here is the usual method being applied for an ellipse:

$x=a \cos t$

$\frac {dx}{dt}=-a\sin t$

$y = b\sin t$

$\frac {dy}{dt}=b\cos t$

$\frac{dy}{dx}=\frac{b \cos t}{-a\sin t}$

Tangent passes through point $(a\cos t,b\sin t)$

$y-b\sin t=-\frac{b \cos t}{a\sin t}(x-a\cos t)$

$ay\sin t-ab\sin^2 t=-bx\cos t+ab\cos^2 t$

$ay\sin t+bx\cos t=ab\cos^2 t+ab\sin^2 t$

$ay\sin t+bx\cos t=ab$

In a similar way we can find the equations of the tangents for the other conics:

Conic	Equation of tangent
Ellipse	$ay\sin t+bx\cos t=ab$
Parabola	$ty-x=at^2$
Hyperbola	$bx \sec t-ay \tan t=ab$

I have noticed an interesting shortcut to finding the equation of a tangent.

I'm applying this shortcut to a hyperbola.

Start with the Cartesian equation: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Perform a partial substitution using the parametric forms: $x=a \sec t, y=b \tan t$. By a partial substitution I mean that I am replacing just one of the $x$ and $y$ with the parametric equivalent.

$\frac{xa \sec t}{a^2}-\frac{yb \tan t}{b^2}=1$

$\frac{x \sec t}{a}-\frac{y \tan t}{b}=1$ $bx \sec t-ay \tan t=ab$

This shortcut works for both the ellipse and the hyperbola.

The parabola needs a slightly different approach.

Rewrite the Cartesian equation $y^2=4ax$ as $y^2=2ax+ 2ax$

Then perform the partial substitution: $y \times 2at=2ax+ 2a \times at^2$

$2aty=2ax+ 2a^2t^2$

$ty=x+ at^2$

$ty-x=at^2$

Neat as the shortcut is, it seems unreasonable that it should work so effectively.

I see that there is another similar question: The Instant Tangent , but I would like to know if there are any other curves where this shortcut can be applied successfully. Or is there something special about the conics that make this shortcut work?

Answer 1

For unified expressions, let $(x_1, y_1)$ be the point on the ellipse $\frac{x^2}{a^2} +\frac{y^2}{b^2}=1$, the hyperbola $\frac{x^2}{a^2} -\frac{y^2}{b^2}=1$, and the parabola $y=ax^2$. Then, their tangent lines are respectively \begin{align} \text{Ellipse:} &\>\>\>\>\> \frac{xx_1}{a^2}+ \frac{yy_1}{b^2}=1\\ \text{Hyperbola:} & \>\>\>\>\> \frac{xx_1}{a^2}- \frac{yy_1}{b^2}=1\\ \text{Parabola:}& \>\>\>\>\> \frac{y+y_1}2={axx_1} \end{align}

The ‘short-cuts’ also work in cases where the point $(x_1, y_1)$ is not on the curve. Then, there are a pair of tangent lines drawn to the curve from the point, which are given by

\begin{align} \text{Ellipse:} &\>\>\>\>\> (\frac{x^2}{a^2} +\frac{y^2}{b^2}-1)(\frac{x_1^2}{a^2} +\frac{y_1^2}{b^2}-1) =(\frac{x_1x}{a^2} +\frac{y_1y}{b^2}-1)^2\\ \text{Hyperbola:} &\>\>\>\>\> (\frac{x^2}{a^2} -\frac{y^2}{b^2}-1)(\frac{x_1^2}{a^2} -\frac{y_1^2}{b^2}-1) =(\frac{x_1x}{a^2} -\frac{y_1y}{b^2}-1)^2\\ \text{Parabola:} &\>\>\>\>\> (y-a x^2)(y_1 -ax_1^2)= \left( \frac{y+y_1}2-axx_1\right)^2 \end{align}