Tangents to Ellipse, Hyperbola and Parabola

Tangent to an Ellipse

A tangent to an ellipse is a straight line, having a single common point with the ellipse.

Consider the canonical equation of an ellipse centered at the origin:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$

Let the tangent be given in parametric form:

$\begin{cases} x = {x_0} + lt \\ y = {y_0} + mt \end{cases},$

where M0(x0, y0) is a point on the ellipse and d(l, m) is the direction vector of the tangent line.

The equation of the tangent line to the ellipse at this point is given by

$\frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1$

Proof

We need to find a unique solution to the system of equations

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\;\;\text{ and }\;\; \begin{cases} x = {x_0} + lt \\ y = {y_0} + mt \end{cases}.$

Substitute the expression for x and y into the equation of the ellipse:

$\frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} + 2t\left({\frac{x_0l}{a^2} + \frac{y_0m}{b^2}}\right) + t^2\left({\frac{l^2}{a^2} + \frac{m^2}{b^2}}\right) = 1.$

Since point M0(x0, y0) belongs to the ellipse, we have

$\frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1,$

that is

$2t\left({\frac{x_0l}{a^2} + \frac{y_0m}{b^2}}\right) + t^2\left({\frac{l^2}{a^2} + \frac{m^2}{b^2}}\right) = 0.$

In addition, point M0(x0, y0) in the equation of the tangent corresponds to t = 0. Therefore the number t = 0 must be the only solution to the last quadratic equation. So we get

$2t\left({\frac{x_0l}{a^2} + \frac{y_0m}{b^2}}\right) + t^2\left({\frac{l^2}{a^2} + \frac{m^2}{b^2}}\right) = 0,$

or

$t\left[{2\left({\frac{x_0l}{a^2} + \frac{y_0m}{b^2}}\right) + t\left({\frac{l^2}{a^2} + \frac{m^2}{b^2}}\right)}\right] = 0.$

Consequently, we arrive at the equation

$\frac{x_0l}{a^2} + \frac{y_0m}{b^2} = 0,$

which establishes the relationship between the components l and m of the direction vector. We can take the following values l and m satisfying this equation:

$l = \frac{y_0}{b^2}, \; m = -\frac{x_0}{a^2}.$

Let's write the tangent equation in the point-direction form:

$\frac{x - x_0}{l} = \frac{y - y_0}{m}, \Rightarrow \frac{x - x_0}{\frac{y_0}{b^2}} = \frac{y - y_0}{-\frac{x_0}{a^2}}, \Rightarrow \left({x - x_0}\right)\frac{x_0}{a^2} + \left({y - y_0}\right)\frac{y_0}{b^2} = 0.$

Simplifying the last equation we get

$\frac{xx_0}{a^2} - \frac{x_0^2}{a^2} + \frac{yy_0}{b^2} - \frac{y_0^2}{b^2} = 0, \Rightarrow \frac{xx_0}{a^2} + \frac{yy_0}{b^2} = \frac{x_0^2}{a^2} + \frac{y_0^2}{b^2}.$

Take into account that the expression on the right side is equal to 1. As a result, we obtain the equation of the tangent to the ellipse in the form

$\frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1.$

Tangent to a Hyperbola

A tangent to a hyperbola is a line that has a single common point with the hyperbola and is not parallel to the asymptotes of the hyperbola.

The equation of a hyperbola centered at the origin with the transverse axis along the x-axis and conjugate axis along the y-axis is given by

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

The equation of a tangent at a point M0(x0, y0) of the hyperbola has the form

$\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1$

Tangent to a Parabola

A tangent to a parabola is a line that has a single point with the parabola and is not parallel to the axis of the parabola.

The equation of a parabola with its vertex at the origin and axis along the x-axis is given by

$y^2 = 2px$

The equation of a tangent at a point M0(x0, y0) of the parabola has the form

$yy_0 = p\left({x + x_0}\right)$

Solved Problems

Example 1.

Find the equation of the tangent to the ellipse $\frac{x^2}{16} + \frac{y^2}{12} = 1$ at the point $$\left({2,3}\right).$$

Solution.

A tangent line to the ellipse is determined by the equation

$\frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1.$

Substitute the coordinates of the tangent point and the parameters of our ellipse:

$\frac{x\cdot2}{16} + \frac{y\cdot3}{12} = 1, \Rightarrow \frac{x}{8} + \frac{y}{4} = 1, \Rightarrow x + 2y = 8.$

So the equation of the tangent line is given by

$x + 2y = 8\;\text{ or }\;x + 2y - 8 = 0.$

Example 2.

Find the equation of the tangent to the hyperbola $\frac{x^2}{32} - \frac{y^2}{8} = 1$ passing through the point $$\left({6,1}\right).$$

Solution.

The equation of a tangent line to the hyperbola is given by

$\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1.$

In our case, the tangency point has coordinates $$\left({6,1}\right).$$ The squares of the semi-axes of the hyperbola are equal $$a^2 = 32$$ and $$b^2 = 8.$$ Then we have

$\frac{x\cdot6}{32} - \frac{y\cdot1}{8} = 1, \Rightarrow \frac{3x}{16} - \frac{y}{8} = 1, \Rightarrow 3x - 2y = 16.$

Therefore the equation of the tangent has the form

$3x - 2y = 16\;\text{ or }\;3x - 2y - 16 = 0.$

Example 3.

Find the equations of the tangents to the ellipse $\frac{x^2}{30} + \frac{y^2}{24} = 1,$ which are parallel to the line $$2x - y - 3 = 0.$$

Solution.

Let's solve this problem using the algebraic method without using the derivative. First we write down the slope $$k$$ of the straight line:

$2x - y - 3 = 0, \Rightarrow y = 2x - 3, \Rightarrow k = 2.$

The tangent to an ellipse is given by the equation

$\frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1.$

Let's express the slope of the tangent from this equation:

$\frac{yy_0}{b^2} = 1 - \frac{xx_0}{a^2}, \Rightarrow y = \frac{b^2}{y_0} - x\frac{b^2x_0}{a^2y_0}, \Rightarrow k = -\frac{b^2x_0}{a^2y_0}.$

Accordingly in our case we get

$k = -\frac{24x_0}{30y_0} = -\frac{4x_0}{5y_0} = 2.$

Therefore, the coordinates of the tangent points satisfy the relation

$x_0 = -\frac{5}{2}y_0.$

Let's substitute this into the equation of the ellipse and find the coordinates of the tangent points:

$\frac{x_0^2}{30} + \frac{y_0^2}{24} = 1, \Rightarrow \frac{\frac{25}{4}y_0^2}{30} + \frac{y_0^2}{24} = 1, \Rightarrow \frac{5y_0^2}{24} + \frac{y_0^2}{24} = 1, \Rightarrow \frac{6y_0^2}{24} = 1, \Rightarrow \frac{y_0^2}{4} = 1, \Rightarrow y_0 = \pm2.$

When $$y_0 = 2,$$ then $$x_0 = -5$$ and when $$y_0 = -2,$$ then $$x_0 = 5.$$ That is, the tangent points have coordinates $${M_1\left({-5,2}\right)}$$ and $${M_2\left({5,-2}\right)}.$$

Finally, we substitute these coordinates into the tangent equation:

$l_1:\;\frac{xx_0}{30} + \frac{yy_0}{24} = 1, \Rightarrow \frac{5x}{30} + \frac{-2y}{24} = 1, \Rightarrow \frac{x}{6} - \frac{y}{12} = 1, \Rightarrow 2x - y - 12 = 0.$
$l_2:\;\frac{xx_0}{30} + \frac{yy_0}{24} = 1, \Rightarrow \frac{-5x}{30} + \frac{2y}{24} = 1, \Rightarrow -\frac{x}{6} + \frac{y}{12} = 1, \Rightarrow 2x - y + 12 = 0.$

Example 4.

Find the equation of the tangent to the parabola $$y^2 = 8x,$$ which is perpendicular to the line $$2x + y - 2 = 0.$$

Solution.

We rewrite the given straight line in the form $$y = -2x + 2,$$ so its slope is equal $$k_1 = -2.$$ By condition, the tangent is perpendicular to the given line. Therefore, the slope of the tangent $$k_2$$ is equal to

$k_1k_2 = -1, \Rightarrow k_2 = -\frac{1}{k_1} = -\frac{1}{-2} = \frac{1}{2}.$

The equation of the tangent to a parabola is given by

$yy_0 = p\left({x + x_0}\right).$

Let us express the slope of the tangent from this formula:

$y = \frac{p}{y_0}x + \frac{px_0}{y_0}, \Rightarrow k_2 = \frac{p}{y_0}.$

Now it's easy to find the $$y-$$coordinate of the tangent point $$M_0:$$

$y_0 = \frac{p}{k_2} = \frac{4}{\frac{1}{2}} = 8.$

then the $$x-$$coordinate of the tangent point is equal to

$x_0 = \frac{y_0^2}{8} = \frac{8^2}{8} = 8.$

Hence, the equation of the tangent to the parabola is given by

$y\cdot8 = 4\left({x + 8}\right), \Rightarrow 2y = x + 8\;\text{ or }\;x - 2y + 8 = 0.$