Question

question 16 · 1 point
what is the standard form equation of the hyperbola that has vertices (0,±1) and foci (0,±5√2)?
select the correct answer below:
○ x²/49 - y² = 1
○ x² - y²/49 = 1
○ y²/49 - x² = 1
○ y² - x²/49 = 1

Explanation:

Step1: Identify the orientation

The vertices are $(0,\pm1)$ and foci are $(0,\pm5\sqrt{2})$. Since the vertices and foci lie on the y - axis, the hyperbola has a vertical transverse axis. The standard form of a hyperbola with a vertical transverse axis is $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}} = 1$.

Step2: Determine the value of $a$

The vertices of a hyperbola with a vertical transverse axis are given by $(0,\pm a)$. Given vertices $(0,\pm1)$, so $a = 1$ and $a^{2}=1$.

Step3: Determine the value of $c$

The foci of a hyperbola with a vertical transverse axis are given by $(0,\pm c)$. Given foci $(0,\pm5\sqrt{2})$, so $c = 5\sqrt{2}$ and $c^{2}=(5\sqrt{2})^{2}=50$.

Step4: Find the value of $b^{2}$

We know the relationship $c^{2}=a^{2}+b^{2}$ for a hyperbola. Substituting $a^{2}=1$ and $c^{2}=50$ into the equation, we get $50 = 1 + b^{2}$. Then $b^{2}=c^{2}-a^{2}=50 - 1=49$.

Step5: Write the equation

Substituting $a^{2}=1$ and $b^{2}=49$ into the standard - form $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}} = 1$, we get $\frac{y^{2}}{1}-\frac{x^{2}}{49}=1$.

Answer:

C. $\frac{y^{2}}{1}-\frac{x^{2}}{49}=1$ (equivalent to $\frac{y^{2}}{49}-x^{2} = 1$ is the correct - form from the given options)