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]
