The lines of the form x cos phi + y sin phi = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The equation of the hyperbola is $4x^2 - y^2 = 4a^2$,which can be written as $\frac{x^2}{a^2} - \frac{y^2}{4a^2} = 1$.Since the chord $x \cos \phi

Vedclass · Accepted Answer

$\frac{2a}{\sqrt{3}}$

Vedclass · Answer

(A) The equation of the hyperbola is $4x^2 - y^2 = 4a^2$,which can be written as $\frac{x^2}{a^2} - \frac{y^2}{4a^2} = 1$.Since the chord $x \cos \phi + y \sin \phi = P$ subtends a right angle at the origin $(0,0)$,we homogenize the equation of the hyperbola using the line equation: $\frac{x^2}{a^2} - \frac{y^2}{4a^2} = \left(\frac{x \cos \phi + y \sin \phi}{P}\right)^2$.Expanding this,we get $x^2 \left(\frac{1}{a^2} - \frac{\cos^2 \phi}{P^2}\right) - y^2 \left(\frac{1}{4a^2} + \frac{\sin^2 \phi}{P^2}\right) - \frac{2xy \cos \phi \sin \phi}{P^2} = 0$.For the chord to subtend a right angle at the origin,the sum of the coefficients of $x^2$ and $y^2$ must be zero:$\left(\frac{1}{a^2} - \frac{\cos^2 \phi}{P^2}\right) - \left(\frac{1}{4a^2} + \frac{\sin^2 \phi}{P^2}\right) = 0$.$\frac{1}{a^2} - \frac{1}{4a^2} - \frac{1}{P^2} (\cos^2 \phi + \sin^2 \phi) = 0$.$\frac{3}{4a^2} - \frac{1}{P^2} = 0$ $\Rightarrow P^2 = \frac{4a^2}{3}$ $\Rightarrow P = \frac{2a}{\sqrt{3}}$.Since $P$ is the perpendicular distance from the origin to the chord,it represents the radius of the circle touched by these chords. Thus,the radius is $\frac{2a}{\sqrt{3}}$.

The lines of the form $x \cos \phi + y \sin \phi = P$ are chords of the hyperbola $4x^2 - y^2 = 4a^2$ which subtend a right angle at the centre of the hyperbola. If these chords touch a circle with centre at $(0,0)$,then the radius of that circle is

Similar Questions

The foci of the hyperbola $4x^2 - 9y^2 - 36 = 0$ are:

Let $P$ be a point on the hyperbola $H: \frac{x^2}{9}-\frac{y^2}{4}=1$,in the first quadrant such that the area of the triangle formed by $P$ and the two foci of $H$ is $2 \sqrt{13}$. Then,the square of the distance of $P$ from the origin is

Find the equation of the hyperbola with eccentricity $e = 3/2$ and foci at $(\pm 2, 0)$.

The eccentricity of the hyperbola conjugate to the hyperbola $\frac{x^2}{4} - \frac{y^2}{12} = 1$ is

Let the foci of a hyperbola be $(1, 14)$ and $(1, -12)$. If it passes through the point $(1, 6)$,then the length of its latus-rectum is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module