Finding the Equation of a Hyperbola Given a Focus and a Tangent Line

To find the equation of a hyperbola when given a focus and a tangent line, follow these steps:

• First, determine whether the transverse axis is along the x-axis $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ or the y-axis $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$$ . This will help you identify the standard form of the hyperbola's equation.
• Using the coordinates of the focus, apply the relationship $c^2 = a^2 + b^2$, where $a$ and $b$ are unknowns.
• Next, compare the equation of the tangent line with the tangent point formula $\frac{x_0x}{a^2} - \frac{y_0y}{b^2} = 1$ to find the coordinates $(x_0; y_0)$ of the point of tangency in terms of $a^2$ and $b^2$. Substitute these into the equation of the tangent line to obtain another equation with $a$ and $b$ as unknowns.
• Finally, set up a system of equations using the two equations you’ve derived and solve for $a^2$ and $b^2$.

By following these steps, you can write the equation of the hyperbola.

Example

Consider a hyperbola centered at the origin with a focus at $F( \sqrt{7},0 )$ and a tangent line given by $x - y = 1$.

The focus is located on the x-axis, so the equation of the hyperbola is:

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

Here, $a$ and $b$ are the semi-axes of the hyperbola. Since the foci are along the x-axis, we use the relationship $c^2 = a^2 + b^2$, where $c$ is the focal distance.

The focus is at $F( \sqrt{7},0 )$, meaning the distance from the center to the focus is $c=\sqrt{7}$.

$$c = \sqrt{7}$$

Knowing that for a hyperbola $c^2 = a^2 + b^2$, we have:

$$c^2 = a^2 + b^2$$

$$(\sqrt{7} )^2 = a^2 + b^2$$

$$a^2 + b^2 = 7 \quad \text{(Equation 1)}$$

Now, we need to find another equation involving $a$ and $b$.

The other piece of information provided is the tangent line, given by the equation $x - y = 1$.

For a hyperbola, the equation of the tangent line at a point $(x_0, y_0)$ on the hyperbola is:

$$\frac{x_0x}{a^2} - \frac{y_0y}{b^2} = 1$$

By comparing this with the equation of the tangent line $x - y = 1$, we get:

$$\underbrace{ \frac{x_0}{a^2} }_1 x - \underbrace{ \frac{y_0}{b^2} }_1 y = 1 \Leftrightarrow x - y = 1$$

This comparison tells us that the coefficients of x and y are:

$$\frac{x_0}{a^2} = 1 \quad \text{and} \quad -\frac{y_0}{b^2} = -1$$

From this, we conclude that

$$x_0 = a^2 \quad \text{and} \quad y_0 = b^2$$

So, the point of tangency $(x_0, y_0)$ on the hyperbola is $(a^2, b^2)$.

Since this point $(x_0, y_0) = (a^2, b^2)$ lies on both the hyperbola and the tangent line $x - y = 1$, we substitute $x = a^2$ and $y = b^2$ into the tangent line equation:

$$a^2 - b^2 = 1 \quad \text{(Equation 2)}$$

This gives us a second equation involving $a$ and $b$.

Now, we can set up a system of equations using the two equations we’ve found and solve the system:

$$\begin{cases} a^2 + b^2 = 7 \\ \\ a^2 - b^2 = 1 \end{cases}$$

We solve for $a^2$ in the first equation and substitute it into the second equation:

$$\begin{cases} a^2 = 7 - b^2 \\ \\ a^2 - b^2 = 1 \end{cases}$$

$$\begin{cases} a^2 = 7 - b^2 \\ \\ 7 - b^2 - b^2 = 1 \end{cases}$$

$$\begin{cases} a^2 = 7 - b^2 \\ \\ -2b^2 = 1 - 7 \end{cases}$$

$$\begin{cases} a^2 = 7 - b^2 \\ \\ -2b^2 = -6 \end{cases}$$

$$\begin{cases} a^2 = 7 - b^2 \\ \\ 2b^2 = 6 \end{cases}$$

$$\begin{cases} a^2 = 7 - b^2 \\ \\ b^2 = \frac{6}{2} \end{cases}$$

$$\begin{cases} a^2 = 7 - b^2 \\ \\ b^2 = 3 \end{cases}$$

After finding $b^2 = 3$, we substitute it back into the first equation to solve for $a^2$:

$$\begin{cases} a^2 = 7 - 3 \\ \\ b^2 = 3 \end{cases}$$

$$\begin{cases} a^2 = 4 \\ \\ b^2 = 3 \end{cases}$$

Now that we know $a^2 = 4$ and $b^2 = 3$, we can substitute these values into the standard equation of the hyperbola:

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

$$\frac{x^2}{4} - \frac{y^2}{3} = 1$$

This is the equation of the hyperbola with a focus at $F( \sqrt{7},0 )$ and a tangent line $x - y = 1$.

And that’s how it’s done.