Question

experiments are performed in which moving particles are deflected by various forces. the particles are eventually deflected along hyperbolic paths. looking at the given figure, if a particle gets as close as 4 units to the nucleus along a hyperbolic path with an asymptote given by $y = \frac{1}{7}x$, what is the equation of its path? the equation of the hyperbolic path is \boxed{ } (type your answer in standard form. use integers or fractions for any numbers in the equation.)

Explanation:

Step1: Identify hyperbola type

The hyperbola opens right, center at (h,k)=(4,0). Standard form: $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2}=1$.

Step2: Find asymptote slope

Asymptote: $y=\frac{1}{7}x$, so $\frac{b}{a}=\frac{1}{7} \Rightarrow b=\frac{a}{7}$.

Step3: Use closest distance

Closest distance from nucleus (origin) to hyperbola is 4. For hyperbola $\frac{(x-4)^2}{a^2}-\frac{y^2}{b^2}=1$, the closest point is (4+a,0) but wait—correction: the distance from origin to the hyperbola's vertex (4+a,0) is not it; actually, the closest distance from origin to the hyperbola is the distance from origin to the vertex (since vertex is (4+a,0)? No, wait the graph shows vertex at (4,0)? No, (4,0) is on x-axis, and the hyperbola passes near it. Wait, the closest distance from nucleus (0,0) to hyperbola is 4. The hyperbola has center at (c,0) where c is distance from center to focus, but wait asymptote $y=\frac{b}{a}(x-h)$ if center is (h,0). Let center be (h,0), asymptote $y=\frac{b}{a}(x-h)=\frac{1}{7}x$. So $\frac{b}{a}(x-h)=\frac{1}{7}x \Rightarrow (\frac{b}{a}-\frac{1}{7})x - \frac{bh}{a}=0$ for all x, so $\frac{b}{a}=\frac{1}{7}$ and $\frac{bh}{a}=0 \Rightarrow h=0$? No, the graph shows nucleus at origin, hyperbola on right. The closest distance from origin to hyperbola is 4, which is the distance from origin to the vertex of the hyperbola. For hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, vertex at (a,0), so distance from origin to vertex is a=4. Then $b=\frac{a}{7}=\frac{4}{7}$.

Step4: Write equation

Substitute a=4, b=4/7 into standard form: $\frac{x^2}{16} - \frac{y^2}{(16/49)}=1 \Rightarrow \frac{x^2}{16} - \frac{49y^2}{16}=1 \Rightarrow x^2 - 49y^2=16$.

Answer:

$x^2 - 49y^2 = 16$