Question

Determine the vertices, asymptotes, and foci of the hyperbola $576x^2 - 16y^2=144$

Please show all your work - thanks! :)​

Answer 1

EXPLANATION.

Equation of the Hyperbola.

⇒ 576x² - 16y² = 144.

As we know that,

Standard equation of Hyperbola = x²/a² - y²/b² = 1.

Compare both the equation, we get.

⇒ 576x²/144 - 16y²/144 = 1.

⇒ x²/144/576 - y²/144/16 = 1.

⇒ x²/1/4 - y²/9 = 1.

⇒ a² = 1/4.

⇒ a = 1/2.

⇒ b² = 9.

⇒ b = 3.

As we know that,

Vertices of Hyperbola = (a,0) and (-a,0).

Vertices of Hyperbola = (1/2,0) and (-1/2,0).

Eccentricity of the Hyperbola = b² = a²(e² - 1).

⇒ 9 = 1/4(e² - 1).

⇒ 36 = e² - 1.

⇒ 37 = e².

⇒ e = √37.

Foci of Hyperbola = S(ae,0) and S'(-ae,0).

Foci of Hyperbola = S(1/2(√37),0) and S'(-(1/2)(√37),0).

Foci of Hyperbola = S(√37/2,0) and S'(-√37/2,0).

Asymptotes of Hyperbola = y = ±b(x)/a.

Asymptotes of Hyperbola = y = ±3x/1/2.

Asymptotes of Hyperbola = y = ±6x.

                                                                                                                         

MORE INFORMATION.

Equation of normal.

(1) = The equation of normal to the hyperbola : x²/a² - y²/b² = 1 at (x₁, y₁) is,

a²x/x₁ + b²y/y₁ = a² + b² = a²e².

(2) = The equation of normal at (a sec∅, b tan∅) to the Hyperbola x²/a² - y²/b² = 1 is ax cos∅ + by cot∅ = a² + b².

(3) = Slope form : y = mx - m(a² + b²)/√a² - b²m².

Answer 2

Answer:

Thevenin's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load.

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