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

Samyak Jain
333 Points
4 years ago
Equation of hyperbola : x/ (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$\dpi{100} \theta$, b tan$\dpi{100} \theta$) i.e. ((1/2)sec$\dpi{100} \theta$ , (1/3)tan$\dpi{100} \theta$)
where $\dpi{100} \theta$ is the eccentric angle of point P.

Let point of contact of tangent 4x – 3y = $\dpi{80} \sqrt{3}$ on the hyperbola be P.
Then the point P ((1/2)sec$\dpi{100} \theta$ , (1/3)tan$\dpi{100} \theta$) satisfies the equation of tangent.

$\dpi{80} \therefore$ 4.(1/2)sec$\dpi{100} \theta$ – 3.(1/3)tan$\dpi{100} \theta$ = $\dpi{80} \sqrt{3}$     $\dpi{100} \Rightarrow$    2 sec$\dpi{100} \theta$ – tan$\dpi{100} \theta$ = $\dpi{80} \sqrt{3}$
or 2 / cos$\dpi{100} \theta$  –  sin$\dpi{100} \theta$ / cos$\dpi{100} \theta$  =  $\dpi{80} \sqrt{3}$      $\dpi{100} \Rightarrow$    (2 – sin$\dpi{100} \theta$) / cos$\dpi{100} \theta$ = $\dpi{80} \sqrt{3}$
i.e. 2 – sin$\dpi{100} \theta$ = $\dpi{80} \sqrt{3}$ cos$\dpi{100} \theta$     $\dpi{100} \Rightarrow$    $\dpi{80} \sqrt{3}$ cos$\dpi{100} \theta$ + sin$\dpi{100} \theta$ = 2     $\dpi{100} \Rightarrow$     ($\dpi{80} \sqrt{3}$ / 2) cos$\dpi{100} \theta$ + (1/2) sin$\dpi{100} \theta$ = 1
i.e. cos($\dpi{100} \pi$/6) cos$\dpi{100} \theta$ + sin($\dpi{100} \pi$/6) sin$\dpi{100} \theta$ = 1        or       cos($\dpi{100} \theta$ – $\dpi{100} \pi$/6) = 1
$\dpi{100} \theta$ – $\dpi{100} \pi$/6  =  2n$\dpi{100} \pi$ , n $\dpi{100} \epsilon$ I.
But eccentric angle of hyperbola lies in the interval [0,2$\dpi{100} \pi$). Thus, n should be 0.
$\dpi{100} \therefore$ $\dpi{100} \theta$ = $\dpi{100} \pi$/6 or 30$\dpi{80} \degree$ .