Why Tangents Are Convenient for These Curves
A circle or ellipse can be described by familiar equations in \(x\) and \(y\), but a parametrization gives a second way to describe each point: both coordinates are functions of a parameter. Differentiating those coordinate functions reveals the direction in which the curve is moving at a given parameter value, and therefore its tangent.
The standard parametrizations also make it possible to locate every horizontal and vertical tangent systematically. The key is to distinguish the derivative of \(y\) with respect to the parameter from the slope of the tangent: a horizontal tangent has zero vertical velocity, while a vertical tangent has zero horizontal velocity.
Standard Parametrizations
Let \(h\) and \(k\) be the coordinates of a center, and let \(a\) and \(b\) be positive real numbers. A parametrized ellipse centered at \((h,k)\), with horizontal semiaxis \(a\) and vertical semiaxis \(b\), is
The parameter \(t\) is real. As \(t\) increases through one full period, from \(0\) to \(2\pi\), the parametrization traces the ellipse once. If \(a=b=r\), the ellipse is a circle of radius \(r\) centered at \((h,k)\).
For the ellipse parametrization, differentiation gives
Because \(a,b>0\), these two derivatives cannot both be zero: that would require both \(\sin t=0\) and \(\cos t=0\), which never happens. Thus every parameter value gives a nonzero tangent direction.
The Slope Formula and the Tangent-Line Equation
At a parameter value where \(x'(t)\ne0\), the parametric slope formula gives
This quotient is defined only when \(\sin t\ne0\). The restriction matters: when \(\sin t=0\), the tangent is vertical rather than a line with a finite slope.
A single tangent-line formula can handle both ordinary and vertical tangents. At \(t=t_0\), the point on the ellipse is \((h+a\cos t_0,k+b\sin t_0)\), and its tangent direction is \((-a\sin t_0,b\cos t_0)\). A normal vector to this direction is \((b\cos t_0,a\sin t_0)\), since their dot product is zero. Therefore the line through the point in the tangent direction satisfies
To check the constant on the right, substitute the point on the ellipse: the left side becomes \(ab\cos^2 t_0+ab\sin^2 t_0=ab\). The equation is valid even when one of its coefficients is zero, so it also covers horizontal and vertical tangents.
Worked Example: A Tangent to a Circle
Worked Example: A Tangent to a Circle
Find the tangent line at \(t=\pi/3\) for the circle \(x(t)=2+3\cos t,\ y(t)=-1+3\sin t\).
First evaluate the trigonometric functions: \(\cos(\pi/3)=1/2\) and \(\sin(\pi/3)=\sqrt{3}/2\). The point is
The derivatives are \(x'(t)=-3\sin t\) and \(y'(t)=3\cos t\). At \(t=\pi/3\),
Since \(x'(\pi/3)\ne0\), the slope is
Using the point and slope, the tangent line is \(y+1-\frac{3\sqrt3}{2}=-\frac{\sqrt3}{3}(x-\frac72)\). Equivalently, the circle tangent formula gives
Check the equation at the point: its left side is \(\frac72+\sqrt3(\frac{3\sqrt3}{2})=\frac72+\frac92=8\), as required. The slope of this line is \(-1/\sqrt3\), agreeing with the derivative calculation.
Finding Horizontal and Vertical Tangents
A horizontal tangent occurs where \(y'(t)=0\) and \(x'(t)\ne0\). For the ellipse, \(b\cos t=0\), so \(\cos t=0\). This happens at \(t=\pi/2+n\pi\), where \(n\) is any integer. At these values, \(\sin t\) is either \(1\) or \(-1\), so \(x'(t)=-a\sin t\ne0\), confirming that the tangent is horizontal.
At \(t=\pi/2+2n\pi\), the point is \((h,k+b)\); at \(t=3\pi/2+2n\pi\), it is \((h,k-b)\). The corresponding horizontal tangent lines are \(y=k+b\) and \(y=k-b\).
A vertical tangent occurs where \(x'(t)=0\) and \(y'(t)\ne0\). Here \(-a\sin t=0\), so \(\sin t=0\), which happens at \(t=n\pi\). At these values, \(\cos t\) is either \(1\) or \(-1\), so \(y'(t)=b\cos t\ne0\). The points are \((h+a,k)\) at \(t=2n\pi\), and \((h-a,k)\) at \(t=(2n+1)\pi\). The vertical tangent lines are \(x=h+a\) and \(x=h-a\).
Worked Example: All Horizontal and Vertical Tangents
Worked Example: All Horizontal and Vertical Tangents
Find the horizontal and vertical tangent lines for \(x(t)=-1+5\cos t,\ y(t)=2+3\sin t\), for \(0\le t<2\pi\).
The derivative pair is
For a horizontal tangent, set \(y'(t)=0\): \(3\cos t=0\), so \(t=\pi/2\) or \(t=3\pi/2\) in the stated interval. At \(t=\pi/2\), the point is \((-1+5(0),2+3(1))=(-1,5)\), and \(x'(\pi/2)=-5\ne0\). Its tangent line is \(y=5\). At \(t=3\pi/2\), the point is \((-1+5(0),2+3(-1))=(-1,-1)\), and \(x'(3\pi/2)=5\ne0\). Its tangent line is \(y=-1\).
For a vertical tangent, set \(x'(t)=0\): \(-5\sin t=0\), so \(t=0\) or \(t=\pi\) in the interval. At \(t=0\), the point is \((-1+5,2)=(4,2)\), and \(y'(0)=3\ne0\); the tangent line is \(x=4\). At \(t=\pi\), the point is \((-1-5,2)=(-6,2)\), and \(y'(\pi)=-3\ne0\); the tangent line is \(x=-6\).
These results also follow from the general tangent equation: at \(t=\pi/2\) and \(3\pi/2\) it reduces to \(y-k=b\) and \(y-k=-b\); at \(t=0\) and \(\pi\) it reduces to \(x-h=a\) and \(x-h=-a\). Thus each listed line passes through its corresponding point and has the required orientation.
Worked Example: A Finite-Slope Tangent to an Ellipse
Worked Example: A Finite-Slope Tangent to an Ellipse
Find the tangent at \(t=\pi/4\) for \(x(t)=1+4\cos t,\ y(t)=-2+2\sin t\), and verify the line.
Since \(\cos(\pi/4)=\sin(\pi/4)=\sqrt2/2\), the point is
The derivatives are \(x'(t)=-4\sin t\) and \(y'(t)=2\cos t\). At \(t=\pi/4\), they are \(-2\sqrt2\) and \(\sqrt2\), respectively. Therefore the slope is
The point-slope form is \(y-(-2+\sqrt2)=-\frac12(x-(1+2\sqrt2))\). From the general tangent equation, the line also satisfies
Substituting the point into the simplified equation gives \((1+2\sqrt2)+2(-2+\sqrt2)+3=4\sqrt2\). Its slope is \(-1/2\), matching the derivative calculation, so the equation passes both checks.
A Common Mistake: Confusing Zero Derivatives with Slope Zero
It is tempting to say that \(x'(t)=0\) means the tangent slope is zero. That reverses the roles of the coordinate derivatives. The slope is \(y'(t)/x'(t)\), not \(x'(t)/y'(t)\); when \(x'(t)=0\) and \(y'(t)\ne0\), the tangent is vertical.
Worked Example: Correcting the Slope Interpretation
For \(x(t)=h+a\cos t,\ y(t)=k+b\sin t\), consider \(t=0\). A mistaken calculation might reason that \(x'(0)=0\), and therefore declare the tangent horizontal. But the actual derivatives are
The derivative direction is \((0,b)\), which points vertically. The point is \((h+a,k)\), so the tangent line is \(x=h+a\), not a horizontal line. The general tangent equation confirms this: at \(t=0\) it becomes \(b(x-h)=ab\), hence \(x=h+a\).
Why the Parametrization Helps
The parametrization provides the point, the tangent direction, and the locations of horizontal and vertical tangents using the same two derivatives. The tangent-line equation is particularly useful because it avoids dividing by \(x'(t)\): it remains valid at the two vertical tangencies as well as at the two horizontal ones.
The circle is the special case \(a=b\). Its slope formula simplifies to \(-\cos t/\sin t\) wherever \(\sin t\ne0\), while the locations of its horizontal and vertical tangents are unchanged. Changing the radius or center changes the points and tangent-line positions, but not the parameter values that produce these tangent orientations.
Check Your Understanding
Use the derivative pair and tangent-line equations to answer the following questions.
- For \(x(t)=3+2\cos t,\ y(t)=1+5\sin t\), what is the tangent slope at \(t=\pi/4\)?
- For the same ellipse, give the parameter values in \(0\le t<2\pi\) and the lines of its horizontal tangents.
- For the same ellipse, give the parameter values in \(0\le t<2\pi\) and the lines of its vertical tangents.
- Why is the slope formula \(y'(t)/x'(t)\) not defined at a vertical tangent?
- Write the general tangent-line equation at parameter \(t_0\) for an ellipse centered at \((h,k)\) with positive semiaxes \(a,b\).