Answer:
2a=1
Step-by-step explanation:
The constant difference for a hyperbola [tex]\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1[/tex] is [tex]2a.[/tex]
1. Since hyperbola has foci (-3.5, 0) and (3.5, 0), then [tex]c=3.5.[/tex] Note that [tex]c=\sqrt{a^2+b^2},[/tex] then
[tex]\sqrt{a^2+b^2}=3.5\Rightarrow a^2+b^2=3.5^2.[/tex]
2. Since point (3.5,24) lies on the hyperbola, then
[tex]\dfrac{3.5^2}{a^2}-\dfrac{24^2}{b^2}=1.[/tex]
3. Solve the system of two equations:
[tex]\left\{\begin{array}{l}a^2+b^2=3.5^2\\\dfrac{3.5^2}{a^2}-\dfrac{24^2}{b^2}=1\end{array}\right.[/tex]
From the 1st equation,
[tex]b^2=3.5^2-a^2,[/tex]
then
[tex]\dfrac{3.5^2}{a^2}-\dfrac{24^2}{3.5^2-a^2}=1,\\ \\\dfrac{3.5^2(3.5^2-a^2)-24^2a^2}{a^2(3.5^2-a^2)}=1,\\ \\3.5^4-3.5^2a^2-24^2a^2=3.5^2a^2-a^4,\\ \\a^4-a^2(2\cdot 3.5^2+24^2)+3.5^4=0,\\ \\a^4-600.5a^2+150.0625=0,\\ \\D=(-600.5)^2-4\cdot 150.0625=360000,\\ \\a^2_{1,2}=\dfrac{600.5\pm 600}{2}=\dfrac{1}{4},600\dfrac{1}{4}.[/tex]
For [tex]a^2=\dfrac{1}{4},\ b^2=3.5^2-0.25=12.[/tex]
For [tex]a^2=600\dfrac{1}{4},\ b^2=3.5^2-600.25<0,[/tex] this is impossible, then [tex]a^2=600\dfrac{1}{4}[/tex] is extra solution.
Hence, [tex]a=\dfrac{1}{2}[/tex] and [tex]2a=1.[/tex]
