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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

Mrunal Sonawane , 5 Years ago
Grade 12
anser 1 Answers
Karanveer

Last Activity: 5 Years ago

Answer is π/6
 
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 = π/6 π/6
Hope it helps.

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