# please answer as fast as possible if 4x-3y =√3 is a tangent to a hyperbola 4x^2 - 9y^2 =1 then the eccentric angle of point of contact isa)45b)30c)60d)90please give solution

Karanveer
33 Points
2 years ago

Step by Step Explanation :
Equation of given hyperbola can be written as
$\frac{x^{2}}{1/4} + \frac{y^{2}}{1/9} = 1$
So
$a^{2} = 1/4$
$a= 1/2$
Similarly $b= 1/3$
General point on ellipse $(1/2 \sec \theta,1/3 \tan \theta)$
Using T=0 on ellipse

We get
$4px-9qy = 1$
Where (p,q) is point of contact
Since it's equation of tangent, so by comparing this with given equation of tangent
$4x-3y = \sqrt{3}$
We get p= $1/\sqrt{3}$

And q = $1/3\sqrt{3}$
Since p = $1/2\sec \theta$
$1/2\sec \theta = 1/\sqrt{3}$
$sec\theta = 2/\sqrt{3}$
So $\Theta = %u03C0/6$ π/6
Hope it helps.