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if 4x-3y = √ 3 is a tangent to 4x^2 - 9y^2 =1 then the eccentric angle of point of contact is please give solution

if 4x-3y =√3 is a tangent to 4x^2 - 9y^2 =1 then the eccentric angle of point of contact is
please give solution

Grade:12

1 Answers

Samyak Jain
333 Points
4 years ago
Equation of hyperbola : x2 / (1/4)  +  y2 / (1/9) = 1.  a2 = 1/4 and b2 = 1/9.
So a = 1/2 and b = 1/3.
 
Parametric coordinates of any point P on the hyperbola is (a sec\theta, b tan\theta) i.e. ((1/2)sec\theta , (1/3)tan\theta)
where \theta is the eccentric angle of point P.
 
Let point of contact of tangent 4x – 3y = \sqrt{3} on the hyperbola be P.
Then the point P ((1/2)sec\theta , (1/3)tan\theta) satisfies the equation of tangent.
 
\therefore 4.(1/2)sec\theta – 3.(1/3)tan\theta = \sqrt{3}     \Rightarrow    2 sec\theta – tan\theta = \sqrt{3}
or 2 / cos\theta  –  sin\theta / cos\theta  =  \sqrt{3}      \Rightarrow    (2 – sin\theta) / cos\theta = \sqrt{3}
i.e. 2 – sin\theta = \sqrt{3} cos\theta     \Rightarrow    \sqrt{3} cos\theta + sin\theta = 2     \Rightarrow     (\sqrt{3} / 2) cos\theta + (1/2) sin\theta = 1
i.e. cos(\pi/6) cos\theta + sin(\pi/6) sin\theta = 1        or       cos(\theta – \pi/6) = 1
\theta – \pi/6  =  2n\pi , n \epsilon I.
But eccentric angle of hyperbola lies in the interval [0,2\pi). Thus, n should be 0.
\therefore \theta = \pi/6 or 30\degree .

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