The eccentricity of the hyperbola $3 x^{2} + 7 xy + 2 y^{2} - 11 x - 7 y + 10 = 0$ is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$\sqrt{\frac{2 \sqrt{2}}{\sqrt{2} + 1}}$

$\sqrt{\frac{2 \sqrt{2}}{\sqrt{2} - 1}}$

$\frac{2 \sqrt{2}}{\sqrt{2} + 1}$

$\sqrt{2}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$3 x^{2} + 7 xy + 2 y^{2} - 11 x - 7 y + 10 = 0$

Equation of pair of asymptotes is $3 x^{2} + 7 xy + 2 y^{2} - 11 x - 7 y + k = 0$

If $θ$ is angle between asymptotes then

$\Rightarrow cosθ = \frac{| a + b |}{\sqrt{{(a - b)}^{2} + 4 h^{2}}} w h e r e a = 3, 2 h = 7, b = 2 \Rightarrow \cos θ = \frac{|5|}{\sqrt{{(1)}^{2} + 49}} \Rightarrow \cos θ = \frac{5}{5 \sqrt{2}} = \frac{1}{\sqrt{2}} \Rightarrow cosθ = \cos 45 \Rightarrow θ = 45^{0} Now θ = 2 \sec^{- 1} (e) \Rightarrow e = \sec (\frac{θ}{2}) \Rightarrow e = \frac{1}{\cos \frac{θ}{2}} \Rightarrow e = \frac{1}{\sqrt{\frac{1 + cosθ}{2}}} = \frac{1}{\sqrt{\frac{1 + \frac{1}{\sqrt{2}}}{2}}} = \sqrt{\frac{2 \sqrt{2}}{\sqrt{2} + 1}}$