Question Statement

Prove that the tangent at any point to the ellipse makes equal angles with the lines joining that point with the foci (the reflection property of the ellipse).

Background and Explanation

To prove this property, we use the standard equation of an ellipse and the concept of slopes. We find the slope of the tangent at a general point $P (x_{1}, y_{1})$ using differentiation and then calculate the angles this tangent makes with the lines connecting $P$ to the two foci, $F$ and $F^{'}$ , using the formula for the angle between two lines.

Solution

Step 1: Find the slope of the tangent line

Consider the standard equation of the ellipse: $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$

Let $P (x_{1}, y_{1})$ be any point on the ellipse. To find the slope of the tangent at $P$ , we differentiate equation (i) with respect to $x$ : $\frac{1}{a ^{2}} (2 x) + \frac{1}{b ^{2}} (2 y) \frac{d y}{d x} \frac{2 y}{b ^{2}} \frac{d y}{d x} \frac{d y}{d x} = 0 = - \frac{2 x}{a ^{2}} = - (\frac{b ^{2}}{a ^{2}} \cdot \frac{x}{y})$

At the specific point $P (x_{1}, y_{1})$ , the slope of the tangent ( $m_{1}$ ) is: $m_{1} = - \frac{b ^{2} x _{1}}{a ^{2} y _{1}}$

Step 2: Find the slopes of the lines joining $P$ to the foci

The foci of the ellipse are $F (c, 0)$ and $F^{'} (- c, 0)$ , where $c^{2} = a^{2} - b^{2}$ .

The slope of the line $\overline{P F^{'}}$ (joining $P$ to the left focus) is: $m_{2} = \frac{y _{1} - 0}{x _{1} - ( - c )} = \frac{y _{1}}{x _{1} + c}$

The slope of the line $\overline{P F}$ (joining $P$ to the right focus) is: $m_{3} = \frac{y _{1} - 0}{x _{1} - c} = \frac{y _{1}}{x _{1} - c}$

Step 3: Calculate the angle $α$ between the tangent and $\overline{P F^{'}}$

Let $α$ be the angle between the tangent line and $\overline{P F^{'}}$ . Using the formula $tan α = \frac{m _{2} - m _{1}}{1 + m _{2} m _{1}}$ :

$tan α = \frac{\frac{y _{1}}{x _{1} + c} - ( - \frac{b ^{2} x _{1}}{a ^{2} y _{1}} )}{1 + ( \frac{y _{1}}{x _{1} + c} ) ( - \frac{b ^{2} x _{1}}{a ^{2} y _{1}} )} = \frac{\frac{a ^{2} y _{1}^{2} + b ^{2} x _{1} ( x _{1} + c )}{a ^{2} y _{1} ( x _{1} + c )}}{\frac{a ^{2} y _{1} ( x _{1} + c ) - b ^{2} x _{1} y _{1}}{a ^{2} y _{1} ( x _{1} + c )}} = \frac{a ^{2} y _{1}^{2} + b ^{2} x _{1}^{2} + b ^{2} c x _{1}}{a ^{2} x _{1} y _{1} + a ^{2} c y _{1} - b ^{2} x _{1} y _{1}} = \frac{( a ^{2} y _{1}^{2} + b ^{2} x _{1}^{2} ) + b ^{2} c x _{1}}{( a ^{2} - b ^{2} ) x _{1} y _{1} + a ^{2} c y _{1}}$

Since $P (x_{1}, y_{1})$ lies on the ellipse, $\frac{x _{1}^{2}}{a ^{2}} + \frac{y _{1}^{2}}{b ^{2}} = 1$ , which simplifies to $a^{2} y_{1}^{2} + b^{2} x_{1}^{2} = a^{2} b^{2}$ . Also, $a^{2} - b^{2} = c^{2}$ . Substituting these:

$tan α tan α = \frac{a ^{2} b ^{2} + b ^{2} c x _{1}}{c ^{2} x _{1} y _{1} + a ^{2} c y _{1}} = \frac{b ^{2} ( a ^{2} + c x _{1} )}{c y _{1} ( c x _{1} + a ^{2} )} = \frac{b ^{2}}{c y _{1}}$

Step 4: Calculate the angle $β$ between the tangent and $\overline{P F}$

Let $β$ be the angle between the tangent line and $\overline{P F}$ . Using the formula $tan β = \frac{m _{1} - m _{3}}{1 + m _{1} m _{3}}$ :

$tan β = \frac{( - \frac{b ^{2} x _{1}}{a ^{2} y _{1}} ) - \frac{y _{1}}{x _{1} - c}}{1 + ( - \frac{b ^{2} x _{1}}{a ^{2} y _{1}} ) ( \frac{y _{1}}{x _{1} - c} )} = \frac{\frac{- b ^{2} x _{1} ( x _{1} - c ) - a ^{2} y _{1}^{2}}{a ^{2} y _{1} ( x _{1} - c )}}{\frac{a ^{2} y _{1} ( x _{1} - c ) - b ^{2} x _{1} y _{1}}{a ^{2} y _{1} ( x _{1} - c )}} = \frac{- b ^{2} x _{1}^{2} + b ^{2} c x _{1} - a ^{2} y _{1}^{2}}{a ^{2} x _{1} y _{1} - a ^{2} c y _{1} - b ^{2} x _{1} y _{1}} = \frac{b ^{2} c x _{1} - ( a ^{2} y _{1}^{2} + b ^{2} x _{1}^{2} )}{( a ^{2} - b ^{2} ) x _{1} y _{1} - a ^{2} c y _{1}}$

Substituting $a^{2} y_{1}^{2} + b^{2} x_{1}^{2} = a^{2} b^{2}$ and $a^{2} - b^{2} = c^{2}$ :

$tan β tan β = \frac{b ^{2} c x _{1} - a ^{2} b ^{2}}{c ^{2} x _{1} y _{1} - a ^{2} c y _{1}} = \frac{b ^{2} ( c x _{1} - a ^{2} )}{c y _{1} ( c x _{1} - a ^{2} )} = \frac{b ^{2}}{c y _{1}}$

Step 5: Conclusion

From equations (ii) and (iii), we see that: $tan α = tan β$ $\Rightarrow α = β$

Thus, the tangent makes equal angles with the lines joining the point of tangency to the foci.

Key Formulas or Methods Used

Equation of Ellipse: $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
Implicit Differentiation: To find the slope of the tangent $\frac{d y}{d x}$ .
Slope Formula: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Angle between two lines: $tan θ = \frac{m _{2} - m _{1}}{1 + m _{1} m _{2}}$
Ellipse Constants: $c^{2} = a^{2} - b^{2}$ and $a^{2} y_{1}^{2} + b^{2} x_{1}^{2} = a^{2} b^{2}$ (for a point on the curve).

Summary of Steps

Differentiate the ellipse equation to find the slope of the tangent $m_{1}$ at $P (x_{1}, y_{1})$ .
Determine the coordinates of the foci $F (c, 0)$ and $F^{'} (- c, 0)$ .
Calculate the slopes $m_{2}$ (for $P F^{'}$ ) and $m_{3}$ (for $P F$ ).
Apply the tangent angle formula to find $tan α$ (angle with $P F^{'}$ ) and $tan β$ (angle with $P F$ ).
Simplify the expressions using the identity $a^{2} y_{1}^{2} + b^{2} x_{1}^{2} = a^{2} b^{2}$ and $a^{2} - b^{2} = c^{2}$ .
Observe that $tan α = tan β$ , proving the angles are equal.

Question Statement

Prove that the tangent at any point to the ellipse makes equal angles with the lines joining that point with the foci (the reflection property of the ellipse).

Background and Explanation

Solution

Step 1: Find the slope of the tangent line

Consider the standard equation of the ellipse: $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$

At the specific point $P (x_{1}, y_{1})$ , the slope of the tangent ( $m_{1}$ ) is: $m_{1} = - \frac{b ^{2} x _{1}}{a ^{2} y _{1}}$

Step 2: Find the slopes of the lines joining $P$ to the foci

The foci of the ellipse are $F (c, 0)$ and $F^{'} (- c, 0)$ , where $c^{2} = a^{2} - b^{2}$ .

The slope of the line $\overline{P F^{'}}$ (joining $P$ to the left focus) is: $m_{2} = \frac{y _{1} - 0}{x _{1} - ( - c )} = \frac{y _{1}}{x _{1} + c}$

The slope of the line $\overline{P F}$ (joining $P$ to the right focus) is: $m_{3} = \frac{y _{1} - 0}{x _{1} - c} = \frac{y _{1}}{x _{1} - c}$

Step 3: Calculate the angle $α$ between the tangent and $\overline{P F^{'}}$

Let $α$ be the angle between the tangent line and $\overline{P F^{'}}$ . Using the formula $tan α = \frac{m _{2} - m _{1}}{1 + m _{2} m _{1}}$ :

$tan α tan α = \frac{a ^{2} b ^{2} + b ^{2} c x _{1}}{c ^{2} x _{1} y _{1} + a ^{2} c y _{1}} = \frac{b ^{2} ( a ^{2} + c x _{1} )}{c y _{1} ( c x _{1} + a ^{2} )} = \frac{b ^{2}}{c y _{1}}$

Step 4: Calculate the angle $β$ between the tangent and $\overline{P F}$

Let $β$ be the angle between the tangent line and $\overline{P F}$ . Using the formula $tan β = \frac{m _{1} - m _{3}}{1 + m _{1} m _{3}}$ :

Substituting $a^{2} y_{1}^{2} + b^{2} x_{1}^{2} = a^{2} b^{2}$ and $a^{2} - b^{2} = c^{2}$ :

$tan β tan β = \frac{b ^{2} c x _{1} - a ^{2} b ^{2}}{c ^{2} x _{1} y _{1} - a ^{2} c y _{1}} = \frac{b ^{2} ( c x _{1} - a ^{2} )}{c y _{1} ( c x _{1} - a ^{2} )} = \frac{b ^{2}}{c y _{1}}$

Step 5: Conclusion

From equations (ii) and (iii), we see that: $tan α = tan β$ $\Rightarrow α = β$

Thus, the tangent makes equal angles with the lines joining the point of tangency to the foci.

Key Formulas or Methods Used

Equation of Ellipse: $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
Implicit Differentiation: To find the slope of the tangent $\frac{d y}{d x}$ .
Slope Formula: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Angle between two lines: $tan θ = \frac{m _{2} - m _{1}}{1 + m _{1} m _{2}}$
Ellipse Constants: $c^{2} = a^{2} - b^{2}$ and $a^{2} y_{1}^{2} + b^{2} x_{1}^{2} = a^{2} b^{2}$ (for a point on the curve).

Summary of Steps

Differentiate the ellipse equation to find the slope of the tangent $m_{1}$ at $P (x_{1}, y_{1})$ .
Determine the coordinates of the foci $F (c, 0)$ and $F^{'} (- c, 0)$ .
Calculate the slopes $m_{2}$ (for $P F^{'}$ ) and $m_{3}$ (for $P F$ ).
Apply the tangent angle formula to find $tan α$ (angle with $P F^{'}$ ) and $tan β$ (angle with $P F$ ).
Simplify the expressions using the identity $a^{2} y_{1}^{2} + b^{2} x_{1}^{2} = a^{2} b^{2}$ and $a^{2} - b^{2} = c^{2}$ .
Observe that $tan α = tan β$ , proving the angles are equal.

7.7 Q-7

Question Statement

Background and Explanation

Solution

Step 1: Find the slope of the tangent line

Step 2: Find the slopes of the lines joining $P$ to the foci

Step 3: Calculate the angle $α$ between the tangent and $\overline{P F^{'}}$

Step 4: Calculate the angle $β$ between the tangent and $\overline{P F}$

Step 5: Conclusion

Key Formulas or Methods Used

Summary of Steps

7.7 Q-7

Question Statement

Background and Explanation

Solution

Step 1: Find the slope of the tangent line

Step 2: Find the slopes of the lines joining $P$ to the foci

Step 3: Calculate the angle $α$ between the tangent and $\overline{P F^{'}}$

Step 4: Calculate the angle $β$ between the tangent and $\overline{P F}$

Step 5: Conclusion

Key Formulas or Methods Used

Summary of Steps