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