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Prove that the tangent at P bisect the angle $S_1PS$ where $S_1$ and $S_2$ are foci of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

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This is the part of the problem Hyperbola $\frac{𝑥^2}{100} − \frac{𝑦^2}{64} = 1$ problem

The concept of the problem is based that the tangent bisect the angle equally

Using internal bisector propety $\frac{S_1P}{S_1T}=\frac{SP}{ST}$ and $S_1P-SP=2a$ and $S_1T+ST=2\sqrt{a^2+b^2}$ where $T$ represent the point where the tangent intersect the x-axis

Not able to proceed further

Samar Imam Zaidi
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2 Answers2

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In your diagram and equations, where it shows $S$, it should be $S_2$, right?

Draw point $P'$ a small distance further up the hyperbola.

According to the locus definition of a hyperbola,

$S_1 P - S_2 P = S_1 P' - S_2 P'$

$\therefore S_1P' - S_1P = S_2P' - S_2P$

Notice that, when $PP'$ is small, $S_1P' - S_1P \approx PP'\cos{(\angle S_1 P' P)}$ (To see this, rotate line segment $S_1 P$ counter-clockwise about $S_1$ until it coincides with line segment $S_1 P'$.)

Similarly, $S_2P' - S_2P \approx PP'\cos{(\angle S_2 P' P)}$.

$\therefore \angle S_1 P' P \approx \angle S_2 P' P$

As $P'$ approaches $P$, the approximation becomes equal, and the line through $P$ and $P'$ becomes the tangent to the hyperbola at $P$

So the tangent at $P$ bisects $\angle S_1 P S_2$.

Dan
  • 22,158
1

The two foci are $S_1$ and $S_2$. Let $P(t)$ be on the hyperbola, where $t$ is a parameter, for example, we can have $P(t) = (a \sec t , b \tan t ) $

From the defining property of the hyperbola,

$ \overline{S_1 P} - \overline{S_2 P} = 2 a = Constant \hspace{25pt} (1) $

But,

$\overline{S_1 P} = \sqrt{ (P - S_1) \cdot (P - S_1 ) } $

and

$\overline{S_2 P} = \sqrt{ (P - S_2) \cdot (P - S_2) } $

Now differentiate $(1)$ with respect to $t$, you obtain

$ \dfrac{ \dot{P} \cdot (P - S_1) }{\overline{S_1 P} } -\dfrac{ \dot{P} \cdot (P - S_2) }{\overline{S_2 P} } = 0 $

which can be written as

$ \dfrac{ \dot{P} \cdot ( S_1 P ) }{\overline{S_1 P} } -\dfrac{ \dot{P} \cdot (S_2 P) }{\overline{S_2 P} } = 0 $

And finally

$ \dfrac{ \dot{P} \cdot ( S_1 P ) }{\overline{S_1 P} } = \dfrac{ \dot{P} \cdot (S_2 P) }{\overline{S_2 P} } $

Divide both sides by $\| \dot{P} \| $

$ \dfrac{ \dot{P} \cdot ( S_1 P ) }{\| \dot{P} \| \overline{S_1 P} } = \dfrac{ \dot{P} \cdot (S_2 P) }{ \| \dot{P} \| \overline{S_2 P} } $

The left hand side is cosine of the angle between $\dot{P}$ and $S_1 P$, and the right hand side is the cosine of the angle between $\dot{P}$ and $S_2 P$.

$\dot{P}$ is a vector pointing tangentially to the hyperbola curve, therefore, the angles it makes with $S_1 P$ and $S_2 P$ are equal, and thus the tangent bisects the angle between them.

Hosam Hajeer
  • 21,978