Name: Equation of Chord of Hyperbola in Parametric Form
Uploaded: 2019-07-19
Duration: 1 min 30 s
Description: How to find the equation of chord of a hyperbola joining two points whose parametric coordinates are known? Check out this video to find out.

Question

If $C$ is the centre of a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, S$ , $S^{'}$ its foci and $P$ a point on it.

Prove that $S P \cdot S^{'} P = C P^{2} - a^{2} + b^{2} .$

Accepted Answer

Let, the point $P$ on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $P (a \sec θ, b \tan θ) .$

We know that the focal distance of a point $(x_{1}, y_{1})$ on the hyperbola are $e x_{1} \pm a .$

Hence, we have

$S P \cdot S^{'} P = (a e \sec θ - a) \times (a e secθ + a)$

$\Rightarrow S P \cdot S^{'} P = a^{2} e^{2} \sec^{2} θ - a^{2}$

$\Rightarrow S P \cdot S^{'} P = a^{2} (1 + \frac{b^{2}}{a^{2}}) \sec^{2} θ - a^{2}$

$\Rightarrow S P \cdot S^{'} P = (a^{2} + b^{2}) \sec^{2} θ - a^{2}$

$\Rightarrow S P \cdot S^{'} P = a^{2} \sec^{2} θ + b^{2} \sec^{2} θ - a^{2}$

Using $1 + \tan^{2} θ = \sec^{2} θ$

$\Rightarrow S P \cdot S^{'} P = a^{2} \sec^{2} θ + b^{2} (1 + \tan^{2} θ) - a^{2}$

$\Rightarrow S P \cdot S^{'} P = a^{2} \sec^{2} θ + b^{2} \tan^{2} θ + b^{2} - a^{2} . . . (1)$

Now, the distance of the point $P (a \sec θ, b \tan θ)$ from the centre $C (0, 0)$ is $C P = \sqrt{a^{2} \sec^{2} θ + b^{2} \tan^{2} θ},$

$\Rightarrow S P \cdot S^{'} P = {(C P)}^{2} + b^{2} - a^{2} .$

If $C$ is the centre of a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, S$ , $S^{'}$ its foci and $P$ a point on it.

Prove that $S P \cdot S^{'} P = C P^{2} - a^{2} + b^{2} .$

Important Questions on Hyperbola

Learn and Prepare for any exam you want !

If C is the centre of a hyperbola x2a2-y2b2=1, S, S' its foci and P a point on it. Prove that SP·S'P=CP2-a2+b2.

Important Questions on Hyperbola

Learn and Prepare for any exam you want !

If $C$ is the centre of a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, S$ , $S^{'}$ its foci and $P$ a point on it.

Prove that $S P \cdot S^{'} P = C P^{2} - a^{2} + b^{2} .$