Question Statement

Prove that the tangent at any point of the hyperbola makes equal angles with the lines joining the point with the foci of the hyperbola.

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Background and Explanation

To solve this problem, we use the standard equation of a hyperbola and the concept of derivatives to find the slope of a tangent line. We also utilize the formula for the angle between two lines based on their slopes and the fundamental relationship between the semi-axes ($a,b$) and the focal distance ($c$) of a hyperbola, where $c^2=a^2+b^2$.

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Solution

Let the equation of the hyperbola be: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\dots (i)$

Let $P(x_1,y_1)$ be any point on the hyperbola. The foci of the hyperbola are located at $F(c,0)$ and $F^{\prime }(-c,0)$.

1. Finding the Slope of the Tangent

To find the slope of the tangent at point $P$, we differentiate equation $(i)$ with respect to $x$:

$\begin{array}{rl}\frac{d}{dx}\left(\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)& =\frac{d}{dx}(1)\\ \frac{1}{a^2}(2x)-\frac{1}{b^2}(2y)\frac{dy}{dx}& =0\\ \frac{y}{b^2}\frac{dy}{dx}& =\frac{x}{a^2}\\ \frac{dy}{dx}& =\frac{b^{2}x}{a^{2}y}\end{array}$

The slope of the tangent line at $P(x_1,y_1)$ is: $m_1=\frac{b^2{x}_1}{a^2{y}_1}$

2. Finding the Slopes of the Focal Radii

The lines joining point $P$ to the foci are $P{F}^{\prime }$ and $PF$.

• Slope of $\overline{P{F}^{\prime }}$ ($m_2$): Using the coordinates $P(x_1,y_1)$ and $F^{\prime }(-c,0)$: $m_2=\frac{y_1-0}{x_1-(-c)}=\frac{y_1}{x_1+c}$

• Slope of $\overline{PF}$ ($m_3$): Using the coordinates $P(x_1,y_1)$ and $F(c,0)$: $m_3=\frac{y_1-0}{x_1-c}=\frac{y_1}{x_1-c}$

3. Calculating the Angle $\alpha$ between $P{F}^{\prime }$ and the Tangent

Let $\alpha$ be the angle between the line $\overline{P{F}^{\prime }}$ and the tangent line. Using the tangent angle formula: $\tan \alpha =\frac{m_1-m_2}{1+m_1{m}_2}$

Substituting the values of $m_1$ and $m_2$: $\begin{array}{rl}\tan \alpha & =\frac{\frac{b^2{x}_1}{a^2{y}_1}-\frac{y_1}{x_1+c}}{1+\left(\frac{b^2{x}_1}{a^2{y}_1}\right)\left(\frac{y_1}{x_1+c}\right)}\\ & =\frac{\frac{b^2{x}_1(x_1+c)-a^2{y}_1^2}{(a^2{y}_1)(x_1+c)}}{\frac{a^2{y}_1(x_1+c)+b^2{x}_1{y}_1}{a^2{y}_1(x_1+c)}}\\ & =\frac{b^2{x}_1^2+b^{2}c{x}_1-a^2{y}_1^2}{a^2{x}_1{y}_1+a^{2}c{y}_1+b^2{x}_1{y}_1}\\ & =\frac{(b^2{x}_1^2-a^2{y}_1^2)+b^{2}c{x}_1}{(a^2+b^2)x_1{y}_1+a^{2}c{y}_1}\end{array}$

Since $P(x_1,y_1)$ lies on the hyperbola, $b^2{x}_1^2-a^2{y}_1^2=a^2{b}^2$. Also, for a hyperbola, $a^2+b^2=c^2$. Substituting these: $\begin{array}{rl}\tan \alpha & =\frac{a^2{b}^2+b^{2}c{x}_1}{c^2{x}_1{y}_1+a^{2}c{y}_1}\\ & =\frac{b^2(a^2+c{x}_1)}{c{y}_1(c{x}_1+a^2)}\\ \tan \alpha & =\frac{b^2}{c{y}_1}\end{array}$

4. Calculating the Angle $\beta$ between $PF$ and the Tangent

Let $\beta$ be the angle between the line $\overline{PF}$ and the tangent line: $\tan \beta =\frac{m_3-m_1}{1+m_3{m}_1}$

Substituting the values of $m_1$ and $m_3$: $\begin{array}{rl}\tan \beta & =\frac{\left(\frac{y_1}{x_1-c}\right)-\left(\frac{b^2{x}_1}{a^2{y}_1}\right)}{1+\left(\frac{y_1}{x_1-c}\right)\left(\frac{b^2{x}_1}{a^2{y}_1}\right)}\\ & =\frac{\frac{a^2{y}_1^2-b^2{x}_1(x_1-c)}{(x_1-c)(a^2{y}_1)}}{\frac{(x_1-c)(a^2{y}_1)+b^2{x}_1{y}_1}{(x_1-c)(a^2{y}_1)}}\\ & =\frac{a^2{y}_1^2-b^2{x}_1^2+b^{2}c{x}_1}{a^2{x}_1{y}_1-a^{2}c{y}_1+b^2{x}_1{y}_1}\\ & =\frac{-(b^2{x}_1^2-a^2{y}_1^2)+b^{2}c{x}_1}{(a^2+b^2)x_1{y}_1-a^{2}c{y}_1}\end{array}$

Applying the same identities ($b^2{x}_1^2-a^2{y}_1^2=a^2{b}^2$ and $a^2+b^2=c^2$): $\begin{array}{rl}\tan \beta & =\frac{-a^2{b}^2+b^{2}c{x}_1}{c^2{x}_1{y}_1-a^{2}c{y}_1}\\ & =\frac{b^2(c{x}_1-a^2)}{c{y}_1(c{x}_1-a^2)}\\ \tan \beta & =\frac{b^2}{c{y}_1}\end{array}$

Conclusion

Since $\tan \alpha =\tan \beta$, it follows that $\alpha =\beta$. Thus, the tangent makes equal angles with the focal radii.

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Key Formulas or Methods Used

• Standard Hyperbola Equation: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
• Hyperbola Identity: $c^2=a^2+b^2$
• Slope of Tangent via Differentiation: $\frac{dy}{dx}=\frac{b^{2}x}{a^{2}y}$
• Angle between two lines: $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$
• Point-on-curve property: $b^2{x}_1^2-a^2{y}_1^2=a^2{b}^2$

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Summary of Steps

1. Define the hyperbola and find the slope of the tangent ($m_1$) at point $P(x_1,y_1)$ using implicit differentiation.
2. Determine the slopes of the lines connecting $P$ to the two foci, $F(c,0)$ and $F^{\prime }(-c,0)$.
3. Apply the tangent angle formula to find the angle $\alpha$ between the tangent and the first focal radius.
4. Apply the tangent angle formula to find the angle $\beta$ between the tangent and the second focal radius.
5. Simplify both expressions using the hyperbola's geometric properties to show that $\tan \alpha =\tan \beta$.