A space probe flies by a planet in an hyperbolic orbit. It reaches the vertex of its orbit at (5,0) and then travels along a 2 path that gets closer and closer to the line y=-x. Write an equation that describes the path of the space probe if the 5 center of its hyperbolic orbit is at (0,0).​

Answer:

[tex]\frac{2x^{2} }{{25} } + \frac{2y^{2}}{{25} } = 1[/tex]

Step-by-step explanation:

Since the required hyperbola has its vertex at (5,0), its transverse axis is on the x-axis and center at (0,0).

So, we use the equation in standard form

[tex]\frac{x^{2} }{a^{2} } + \frac{y^{2} }{b^{2} } = 1[/tex]

Also, since the path of the space probe gets closer to y = -x, this is the asymptote to the hyperbola.

Our standard asymptote equation is y = ±bx/a taking the negative sign and comparing with y = -x,

-bx/a = -x ⇒ b/a = 1 ⇒ a = b

Also, the coordinate of the vertex (c, 0) = (5, 0) and c² = a² + b²

substituting c = 5 and a = b into the equation, we have

c² = a² + b²

5² = a² + a²

25 = 2a²

a² = 25/2

a = √(25/2)

a = ±5/√2

rationalizing, we have

a = ±5/√2 × √2/√2

a = ±5√2/2

Since a = b, b = ±5√2/2

Inserting a and b into the equation for the hyperbola, we have

[tex]\frac{x^{2} }{a^{2} } + \frac{y^{2} }{b^{2} } = 1\\\frac{x^{2} }{(\frac{5\sqrt{2} }{2} )^{2} } + \frac{y^{2} }{\frac{5\sqrt{2} }{2} ^{2} } = 1\\\frac{x^{2} }{\frac{25 X 2 }{4} } + \frac{y^{2} }{\frac{25X2 }{4} } = 1\\\frac{x^{2} }{\frac{25}{2} } + \frac{y^{2} }{\frac{25}{2} } = 1\\\frac{2x^{2} }{{25} } + \frac{2y^{2}}{{25} } = 1[/tex]

So, the required equation is

[tex]\frac{2x^{2} }{{25} } + \frac{2y^{2}}{{25} } = 1[/tex]

What is the measure of arc AB?
A. 153
B.140
C.142
D.180

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