Answer to Question #87969 – Math – Analytic Geometry
Question
Find the vertices, eccentricity, foc and asymptotes of the hyperbola x 2 / 8 − y 2 / 4 = 1 x^2/8-y^2/4=1 x 2 /8 − y 2 /4 = 1 ≻ \succ ≻ x + ≻ y = 2 x+\succ y=2 x + ≻ y = 2
Solution
x 2 8 − y 2 4 = 1 \frac{x^2}{8} - \frac{y^2}{4} = 1 8 x 2 − 4 y 2 = 1
Rewrite in the form of a a a
( x − 0 ) 2 ( 2 2 ) 2 − ( y − 0 ) 2 2 2 = 1 \frac{(x - 0)^2}{(2\sqrt{2})^2} - \frac{(y - 0)^2}{2^2} = 1 ( 2 2 ) 2 ( x − 0 ) 2 − 2 2 ( y − 0 ) 2 = 1 ∴ ( h , k ) = ( 0 , 0 ) , a = 2 2 , b = 2 \therefore (h, k) = (0, 0), \, a = 2\sqrt{2}, \, b = 2 ∴ ( h , k ) = ( 0 , 0 ) , a = 2 2 , b = 2 Vertices = ( 0 , 0 ) \text{Vertices} = (0, 0) Vertices = ( 0 , 0 )
The eccentricity = e a 2 + b 2 a = e \frac{\sqrt{a^2 + b^2}}{a} = e a a 2 + b 2
e = ( 2 2 ) 2 + 2 2 2 2 = 3 2 e = \frac{\sqrt{(2\sqrt{2})^2 + 2^2}}{2\sqrt{2}} = \sqrt{\frac{3}{2}} e = 2 2 ( 2 2 ) 2 + 2 2 = 2 3
For right-left hyperbola the asymptotes are y = ± b a ( x − h ) + k y = \pm \frac{b}{a} (x - h) + k y = ± a b ( x − h ) + k
y = 2 2 2 ( x − 0 ) + 0 , y = − 2 2 2 ( x − 0 ) + 0 y = \frac{2}{2\sqrt{2}} (x - 0) + 0, \quad y = -\frac{2}{2\sqrt{2}} (x - 0) + 0 y = 2 2 2 ( x − 0 ) + 0 , y = − 2 2 2 ( x − 0 ) + 0 y = x 2 , y = − x 2 y = \frac{x}{\sqrt{2}}, \quad y = -\frac{x}{\sqrt{2}} y = 2 x , y = − 2 x
The Foci are ( h + c , k ) , ( h − c , k ) (h + c, k), (h - c, k) ( h + c , k ) , ( h − c , k ) c = a 2 + b 2 c = \sqrt{a^2 + b^2} c = a 2 + b 2
c = ( 2 2 ) 2 + 2 2 = 2 3 c = \sqrt{(2\sqrt{2})^2 + 2^2} = 2\sqrt{3} c = ( 2 2 ) 2 + 2 2 = 2 3
Then, foci = ( 0 + 2 3 , 0 ) , ( 0 − 2 3 , 0 ) = (0 + 2\sqrt{3}, 0), (0 - 2\sqrt{3}, 0) = ( 0 + 2 3 , 0 ) , ( 0 − 2 3 , 0 )
= ( 2 3 , 0 ) , ( − 2 3 , 0 ) = (2\sqrt{3}, 0), (-2\sqrt{3}, 0) = ( 2 3 , 0 ) , ( − 2 3 , 0 )
Slope of tangent to curve = d y d x = \frac{dy}{dx} = d x d y
On differentiating,
x 2 8 − y 2 4 = 1 … ( i ) \frac{x^2}{8} - \frac{y^2}{4} = 1 \dots (i) 8 x 2 − 4 y 2 = 1 … ( i ) 2 x 8 − 2 y 4 d y d x = 0 \frac{2x}{8} - \frac{2y}{4} \frac{dy}{dx} = 0 8 2 x − 4 2 y d x d y = 0 d y d x = − 2 x 8 − 2 y 4 = x 2 y \frac{dy}{dx} = -\frac{\frac{2x}{8}}{-\frac{2y}{4}} = \frac{x}{2y} d x d y = − − 4 2 y 8 2 x = 2 y x
From (i), y = ± 4 ( x 2 8 − 1 ) y = \pm \sqrt{4\left(\frac{x^2}{8} - 1\right)} y = ± 4 ( 8 x 2 − 1 )
y = ± x 2 2 − 4 y = \pm \sqrt{\frac{x^2}{2} - 4} y = ± 2 x 2 − 4
so, d y d x = x ± 2 x 2 2 − 4 \frac{dy}{dx} = \frac{x}{\pm 2\sqrt{\frac{x^2}{2} - 4}} d x d y = ± 2 2 x 2 − 4 x
Given line: x + λ y = 2 x + \lambda y = 2 x + λ y = 2
y = − x λ + 2 λ y = \frac{-x}{\lambda} + \frac{2}{\lambda} y = λ − x + λ 2
Slope = − 1 λ = \frac{-1}{\lambda} = λ − 1
We get,
x ± 2 x 2 2 − 4 = − 1 λ \frac{x}{\pm 2\sqrt{\frac{x^2}{2} - 4}} = \frac{-1}{\lambda} ± 2 2 x 2 − 4 x = λ − 1
Thus, λ = ∓ 2 x 2 2 − 4 x \lambda = \mp \frac{2\sqrt{\frac{x^2}{2} - 4}}{x} λ = ∓ x 2 2 x 2 − 4
Applying these elements and tracing the hyperbola, we get following graph:
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