Condition of a Line to be Tangent to an Ellipse

An ellipse is a set of points in a plane, the sum of whose distances from two fixed points (the foci) is constant. The standard form of an ellipse centered at the origin with its major axis along the x-axis is given by the equation:

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

where ( a ) is the semi-major axis and ( b ) is the semi-minor axis.

Tangent Line to an Ellipse

A line is said to be tangent to an ellipse if it touches the ellipse at exactly one point. This point is called the point of tangency. The condition for a line to be tangent to an ellipse can be derived using the concept of the slope or by using the distance from the center to the line.

Slope Form

A line with slope ( m ) and y-intercept ( c ) can be written as:

$$ y = mx + c $$

For this line to be tangent to the ellipse, it must satisfy the ellipse equation at exactly one point. By substituting ( y ) from the line equation into the ellipse equation, we get:

$$ \frac{x^2}{a^2} + \frac{(mx + c)^2}{b^2} = 1 $$

This is a quadratic equation in ( x ). For the line to be tangent to the ellipse, this quadratic equation must have exactly one solution. This happens when the discriminant of the quadratic equation is zero. The discriminant ( D ) is given by:

$$ D = b^2m^2 - 4\left(\frac{c^2}{b^2} - 1\right)a^2 $$

Setting ( D = 0 ), we get the condition for tangency:

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

Point-Slope Form

If the line passes through a point ( (x_1, y_1) ) and has a slope ( m ), the equation of the line is:

$$ y - y_1 = m(x - x_1) $$

For this line to be tangent to the ellipse, the point ( (x_1, y_1) ) must lie on the ellipse, and the slope ( m ) must satisfy the tangency condition derived above.

Perpendicular Distance from Center

Another way to derive the condition for tangency is by using the perpendicular distance from the center of the ellipse to the line. If ( L ) is the line given by:

$$ Ax + By + C = 0 $$

The perpendicular distance ( d ) from the center of the ellipse (which is the origin in this case) to the line ( L ) is given by:

$$ d = \frac{|C|}{\sqrt{A^2 + B^2}} $$

For the line to be tangent to the ellipse, this distance must be equal to the length of the semi-minor axis ( b ) if ( A = 0 ) or the semi-major axis ( a ) if ( B = 0 ). Therefore, the condition for tangency is:

$$ C^2 = a^2B^2 + b^2A^2 $$

Table of Differences and Important Points

Aspect	Slope Form Condition	Perpendicular Distance Condition
Equation of the Line	( y = mx + c )	( Ax + By + C = 0 )
Condition for Tangency	( c^2 = a^2m^2 + b^2 )	( C^2 = a^2B^2 + b^2A^2 )
Discriminant Method	Discriminant ( D ) must be zero	Not applicable
Distance Method	Not directly used	Perpendicular distance ( d )
Point of Tangency	Must satisfy both line and ellipse	Must satisfy both line and ellipse

Examples

Example 1: Using Slope Form

Given an ellipse with ( a = 5 ) and ( b = 3 ), find the equation of the tangent line with slope ( m = 2 ).

Using the slope form condition:

$$ c^2 = a^2m^2 + b^2 $$ $$ c^2 = 5^2 \cdot 2^2 + 3^2 $$ $$ c^2 = 100 + 9 $$ $$ c^2 = 109 $$ $$ c = \pm\sqrt{109} $$

Thus, the equations of the tangent lines are:

$$ y = 2x + \sqrt{109} $$ $$ y = 2x - \sqrt{109} $$

Example 2: Using Perpendicular Distance

Given an ellipse with ( a = 4 ) and ( b = 2 ), find the equation of the tangent line with ( A = 1 ) and ( B = 2 ).

Using the perpendicular distance condition:

$$ C^2 = a^2B^2 + b^2A^2 $$ $$ C^2 = 4^2 \cdot 2^2 + 2^2 \cdot 1^2 $$ $$ C^2 = 64 \cdot 4 + 4 $$ $$ C^2 = 256 + 4 $$ $$ C^2 = 260 $$ $$ C = \pm\sqrt{260} $$

Thus, the equations of the tangent lines are:

$$ x + 2y + \sqrt{260} = 0 $$ $$ x + 2y - \sqrt{260} = 0 $$

Understanding the condition for a line to be tangent to an ellipse is crucial for solving problems related to tangents, normals, and geometric properties of ellipses. It is also a fundamental concept in calculus and analytical geometry.