Question

eccentricity of the hyperbola is 2x^2-5xy+2y^2+10=10​

Answer 1

we have to find the eccentricity of the hyperbola is 2x² - 5xy + 2y² + 10 = 0

solution : 2x² - 5xy + 2y² + 10 = 0

⇒x² - 5xy/2 + y² + 5 = 0

⇒x² - 2(5y/4)x + (5y/4)² - (5y/4)² + y² + 5 = 0

⇒(x - 5y/4)² - 25y²/16 + y² + 5 = 0

⇒(4x - 5y)²/16 - 25y²/16 + 16y²/16 + 80/16 = 0

⇒(4x - 5y)² - 9y² + 80 = 0

⇒(4x - 5y)² - 9y² = -80

⇒(4x - 5y)²/9 - y² = -80

⇒y² - (4x - 5y)²/9 = 80

⇒y²/80 - (4x - 5y)²/720 = 1

⇒y²/(4√5)² - (4x - 5y)²/(12√5)² = 1

here a = 12√5 and b = 4√5

so, eccentricity = √(a² + b²)/a

= √{720 + 80}/12√5

= √(800)/12√5

= (20√2)/(12√5)

= (5√2/3√5)

= √10/3

Therefore the eccentricity of the hyperbola is √10/3