Point of Intersection of Tangents on an Ellipse

In this topic, we will discuss the concept of the point of intersection of tangents on an ellipse. An ellipse is a closed curve that is formed by the intersection of a cone and a plane. It is defined as the set of all points in a plane, the sum of whose distances from two fixed points (called the foci) is constant.

When we draw tangents to an ellipse from an external point, these tangents intersect at a point on the ellipse. This point of intersection is known as the point of intersection of tangents.

Table of Contents

1. Equation of an Ellipse
2. Tangents to an Ellipse
3. Point of Intersection of Tangents
4. Examples

1. Equation of an Ellipse

The equation of an ellipse with its center at the origin (0,0) is given by:

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

where 'a' is the semi-major axis and 'b' is the semi-minor axis of the ellipse.

2. Tangents to an Ellipse

A tangent to an ellipse is a straight line that touches the ellipse at only one point. The slope of the tangent at a given point (x, y) on the ellipse is given by:

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

3. Point of Intersection of Tangents

When we draw tangents to an ellipse from an external point (h, k), these tangents intersect at a point on the ellipse. This point of intersection is known as the point of intersection of tangents.

To find the coordinates of the point of intersection of tangents, we can solve the equations of the tangents with the equation of the ellipse simultaneously.

Let's consider an ellipse with its center at the origin (0,0) and equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. We want to find the point of intersection of tangents from an external point (h, k).

Step 1: Find the slope of the tangent at the point (h, k) using the formula:

$$m = -\frac{b^2h}{a^2k}$$

Step 2: Write the equation of the tangent using the point-slope form:

$$y - k = m(x - h)$$

Step 3: Substitute the equation of the tangent into the equation of the ellipse:

$$\frac{x^2}{a^2} + \frac{(m(x - h) + k)^2}{b^2} = 1$$

Step 4: Simplify the equation and solve for x:

$$x^2b^2 + (m^2a^2)x^2 - 2m^2a^2hx + m^2a^2h^2 + b^2k^2 - b^2a^2 = 0$$

This is a quadratic equation in x. Solve it to find the values of x.

Step 5: Substitute the values of x into the equation of the tangent to find the corresponding y-coordinates:

$$y = m(x - h) + k$$

The coordinates of the point of intersection of tangents are (x, y).

4. Examples

Example 1:

Find the point of intersection of tangents to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ from the external point (5, 2).

Solution: Step 1: Find the slope of the tangent at the point (5, 2) using the formula:

$$m = -\frac{4^2 \cdot 5}{9^2 \cdot 2} = -\frac{40}{324}$$

Step 2: Write the equation of the tangent using the point-slope form:

$$y - 2 = -\frac{40}{324}(x - 5)$$

Step 3: Substitute the equation of the tangent into the equation of the ellipse:

$$\frac{x^2}{9} + \frac{(-\frac{40}{324}(x - 5) + 2)^2}{4} = 1$$

Step 4: Simplify the equation and solve for x:

$$324x^2 + 40^2(x - 5)^2 - 9^2(x - 5)^2 - 4 \cdot 9^2 = 0$$

Solving this quadratic equation, we get x = 5 and x = -5.

Step 5: Substitute the values of x into the equation of the tangent to find the corresponding y-coordinates:

For x = 5, y = -\frac{40}{324}(5 - 5) + 2 = 2

For x = -5, y = -\frac{40}{324}(-5 - 5) + 2 = 2

Therefore, the point of intersection of tangents is (5, 2).

Example 2:

Find the point of intersection of tangents to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ from the external point (3, -4).

Solution: Step 1: Find the slope of the tangent at the point (3, -4) using the formula:

$$m = -\frac{9^2 \cdot 3}{16^2 \cdot -4} = \frac{243}{256}$$

Step 2: Write the equation of the tangent using the point-slope form:

$$y + 4 = \frac{243}{256}(x - 3)$$

Step 3: Substitute the equation of the tangent into the equation of the ellipse:

$$\frac{x^2}{16} + \frac{(\frac{243}{256}(x - 3) - 4)^2}{9} = 1$$

Step 4: Simplify the equation and solve for x:

$$256x^2 + 243^2(x - 3)^2 - 16^2(x - 3)^2 - 9 \cdot 16^2 = 0$$

Solving this quadratic equation, we get x = 3 and x = -3.

Step 5: Substitute the values of x into the equation of the tangent to find the corresponding y-coordinates:

For x = 3, y = \frac{243}{256}(3 - 3) - 4 = -4

For x = -3, y = \frac{243}{256}(-3 - 3) - 4 = -4

Therefore, the point of intersection of tangents is (3, -4).

In conclusion, the point of intersection of tangents on an ellipse can be found by solving the equations of the tangents with the equation of the ellipse simultaneously.