# 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:

Let the tangent be given in parametric form:

where *M*_{0}(*x*_{0}, *y*_{0}) 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

### Proof

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

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

Since point *M*_{0}(*x*_{0}, *y*_{0}) belongs to the ellipse, we have

that is

In addition, point *M*_{0}(*x*_{0}, *y*_{0}) 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

or

Consequently, we arrive at the equation

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:

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

Simplifying the last equation we get

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

## 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

The equation of a tangent at a point *M*_{0}(*x*_{0}, *y*_{0}) of the hyperbola has the form

## 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

The equation of a tangent at a point *M*_{0}(*x*_{0}, *y*_{0}) of the parabola has the form

## 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

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

So the equation of the tangent line is given by

### 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

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

Therefore the equation of the tangent has the form

### 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:

The tangent to an ellipse is given by the equation

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

Accordingly in our case we get

Therefore, the coordinates of the tangent points satisfy the relation

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

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:

### 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

The equation of the tangent to a parabola is given by

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

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

then the \(x-\)coordinate of the tangent point is equal to

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