If P(√2 sec θ, √2 tan θ) is a point on the hyperbola whose distance from the origin is √6 where P is in the first quadrant then θ=

Question

If P(√2 sec θ, √2 tan θ) is a point on the hyperbola whose distance from the origin is √6 where P is in the first quadrant then θ= (A) (π/4) (B) (π/3) (C) (π/6) (D) None of these

Tardigrade Admin · Accepted Answer

√22 sec 2 θ+√22 tan 2 θ=6 ⇒ 1+2 tan 2 θ=3 ∴ θ=π / 4 for first quadrant

Q. If $P (2 sec θ, 2 tan θ)$ is a point on the hyperbola whose distance from the origin is $6$ where $P$ is in the first quadrant then $θ =$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{π}{6}$

None of these

$2^{2} sec^{2} θ + 2^{2} tan^{2} θ = 6 \Rightarrow 1 + 2 tan^{2} θ = 3$
$∴ θ = π /4$ for first quadrant

Q. If P(2​secθ,2​tanθ) is a point on the hyperbola whose distance from the origin is 6​ where P is in the first quadrant then θ=

4π​

3π​

6π​