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

The tangent & normal at a point on

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

cut the

y

-axis respectively at

A

&

B .

Prove that the circle on

A B

as diameter passes through the focii of the hyperbola.

Accepted Answer

Given equation of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ .

Focus $= \pm a e, 0, = (\pm a \sqrt{\frac{a^{2} + b^{2}}{a^{2}}}, 0)$

Focus $= (\pm \sqrt{a^{2} + b^{2}}, 0)$

Now equation of tangent to the hyperbola at any point $(a s e c θ, b \tan θ)$

$= \frac{x \sec θ}{a} - \frac{y \tan θ}{b} = 1 . . . (1)$

Equation $(1)$ cuts the $y$ -axis, so $x = 0$ .

So, $- \frac{y \tan θ}{b} = 1 \Rightarrow y = - \frac{b}{\tan θ}$

So, $A = (0, - \frac{b}{\tan θ})$

Again the equation of normal to the hyperbola the point $(a \sec θ, b \tan θ)$ .

$\Rightarrow a x \cos θ + b y \cot θ = a^{2} + b^{2} . . . (2)$

This equation cuts at $y$ -axis, so $x = 0$ .

So, by $\cot θ = a^{2} + b^{2}$

$y = \frac{a^{2} + b^{2}}{b \cot θ}$

So, $B = (0, \frac{a^{2} + b^{2}}{b \cot θ})$

Now given in the question.

$A B$ is the diameter of the circle.

From any point $a \sec θ, b \tan θ$ .

On the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ the equation of tangent $\frac{x \cos θ}{a} - \frac{y \tan θ}{b} = 1$ and equation of normal $a x \cos θ + b y \cot θ = a^{2} + b^{2}$

So, equation of circle in diameter form.

$(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$

$(x - 0) (x - 0) + (y - \frac{a^{2} + b^{2}}{b \cot θ}) (y + \frac{b}{\tan θ}) = 0$

So, $x^{2} + y^{2} + y (\frac{b}{\tan θ} - \frac{a^{2} + b^{2}}{b \cot θ}) - (a^{2} + b^{2}) = 0 . . . (3)$

So, as asked in the question the circle passing through focus so equation $(3)$ will satisfy the focus of hyperbola.

$a^{2} + b^{2} + 0 + 0 (\frac{b}{\tan θ} - \frac{a^{2} + b^{2}}{b \cot θ}) - (a^{2} + b^{2}) = 0$

$\Rightarrow (a^{2} + b^{2}) - (a^{2} + b^{2}) = 0$

$\Rightarrow 0 = 0$

So, the circle passes through the focus of the hyperbola.

The tangent & normal at a point on $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ cut the $y$ -axis respectively at $A$ & $B .$ Prove that the circle on $A B$ as diameter passes through the focii of the hyperbola.

Important Questions on Hyperbola

Learn and Prepare for any exam you want !

The tangent & normal at a point on x2a2-y2b2=1 cut the y-axis respectively at A & B. Prove that the circle on AB as diameter passes through the focii of the hyperbola.

Important Questions on Hyperbola

Learn and Prepare for any exam you want !

The tangent & normal at a point on $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ cut the $y$ -axis respectively at $A$ & $B .$ Prove that the circle on $A B$ as diameter passes through the focii of the hyperbola.