Question

Write the equation of the hyperbola with center (1,-3), vertex (1,2), and focus (1,-3+2sqrt10)

Answer 1

option c

Explanation:

Answer 2

`\textsf{b)} \quad ((y+3)^2)/(25)-((x-1)^2)/(15)=1`

Explanation:

`\boxed{\begin{minipage}{7.4 cm}\underline{Standard equation of vertical hyperbola}\\\\$((y-k)^2)/(a^2)-((x-h)^2)/(b^2)=1$\\\\where:\\\phantom{ww}$\bullet$ $(h,k)$ is the center. \\ \phantom{ww}$\bullet$ $(h, k\pm a)$ are the vertices. \\\phantom{ww}$\bullet$ $(h\pm b, k)$ are the co-vertices. \\\phantom{ww}$\bullet$ $(h, k\pm c)$ are the foci where $c^2=a^2-b^2$.\\\end{minipage}}`

Given:

• center (1, -3)
• vertex (1, 2)
• focus (1, -3+2√10)

As the x-values of the given center, vertex and focus are the same, the hyperbola is vertical (opening up and down).

Given the center is (1, -3):

• h = 1
• k = -3

Use the vertices formula (h, k±a) to find a:

`\begin{aligned} (h, k \pm a) = (1, -3 \pm a) &= (1, 2)\\\implies -3 \pm a &= 2\\a &= 5\end{aligned}`

Use the foci formula (h, k±c) to find c:

`\begin{aligned} (h, k \pm c) = (1, -3 \pm c)&=(1, -3 \pm 2√(10))\\c &= 2 √(10)\end{aligned}`

Use the found values of a and c with Pythagoras Theorem to find b²:

`\begin{aligned}c^2 &=a^2+b^2\\\implies (2 √(10))^2&=5^2+b^2\\40&=25+b^2\\b^2&=15\end{aligned}`

Therefore:

• h = 1
• k = -3
• a² = 5² = 25
• b² = 15

Substitute the found values into the hyperbola formula to create an equation of the hyperbola with the given parameters:

`((y+3)^2)/(25)-((x-1)^2)/(15)=1`