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assignment 7.5: hyperbolas
score: 6/15 answered: 6/15
question 7
write the equation of the hyperbola centered at the origin, with length of the horizontal transverse axis 2 and the curve passes through the point (7, -3).

Explanation:

Step1: Recall hyperbola standard form

For a hyperbola centered at the origin with a horizontal transverse axis, the standard equation is:
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

Step2: Calculate $a$ from axis length

The length of the horizontal transverse axis is $2a$. Given the length is 2:
$$2a = 2 \implies a = 1 \implies a^2 = 1$$

Step3: Substitute point to find $b^2$

Substitute $x=7$, $y=-3$, and $a^2=1$ into the standard equation:
$$\frac{7^2}{1} - \frac{(-3)^2}{b^2} = 1$$
Simplify and solve for $b^2$:
$$49 - \frac{9}{b^2} = 1$$
$$\frac{9}{b^2} = 49 - 1 = 48$$
$$b^2 = \frac{9}{48} = \frac{3}{16}$$

Step4: Plug values into standard form

Substitute $a^2=1$ and $b^2=\frac{3}{16}$ into the equation:
$$x^2 - \frac{y^2}{\frac{3}{16}} = 1$$
Rewrite the second term for clarity:
$$x^2 - \frac{16y^2}{3} = 1$$

Answer:

$\boldsymbol{x^2 - \frac{16y^2}{3} = 1}$ (or equivalent form $\boldsymbol{\frac{x^2}{1} - \frac{y^2}{\frac{3}{16}} = 1}$)