Question

Find the constant difference for a hyperbola with foci (-3.5, 0) and (3.5, 0) and a point on the hyperbola (3. 5, 24).

Answer 1

Answer:

2a=1

Step-by-step explanation:

The constant difference for a hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ is  $2a.$

1. Since hyperbola has foci (-3.5, 0) and (3.5, 0), then $c=3.5.$ Note that $c=\sqrt{a^2+b^2},$ then

$\sqrt{a^2+b^2}=3.5\Rightarrow a^2+b^2=3.5^2.$

2. Since point (3.5,24) lies on the hyperbola, then

$\dfrac{3.5^2}{a^2}-\dfrac{24^2}{b^2}=1.$

3. Solve the system of two equations:

$\left\{\begin{array}{l}a^2+b^2=3.5^2\\\dfrac{3.5^2}{a^2}-\dfrac{24^2}{b^2}=1\end{array}\right.$

From the 1st equation,

$b^2=3.5^2-a^2,$

then

$\dfrac{3.5^2}{a^2}-\dfrac{24^2}{3.5^2-a^2}=1,\\ \\\dfrac{3.5^2(3.5^2-a^2)-24^2a^2}{a^2(3.5^2-a^2)}=1,\\ \\3.5^4-3.5^2a^2-24^2a^2=3.5^2a^2-a^4,\\ \\a^4-a^2(2\cdot 3.5^2+24^2)+3.5^4=0,\\ \\a^4-600.5a^2+150.0625=0,\\ \\D=(-600.5)^2-4\cdot 150.0625=360000,\\ \\a^2_{1,2}=\dfrac{600.5\pm 600}{2}=\dfrac{1}{4},600\dfrac{1}{4}.$

For $a^2=\dfrac{1}{4},\ b^2=3.5^2-0.25=12.$

For $a^2=600\dfrac{1}{4},\ b^2=3.5^2-600.25$ this is impossible, then $a^2=600\dfrac{1}{4}$ is extra solution.

Hence, $a=\dfrac{1}{2}$ and $2a=1.$