# Hallo, please I assist me understand? (d) Derive the standard equation of the hyperbola with foci along the y-axis and center at (h; k).

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

2

### This answer has been featured!

Featured answers represent the very best answers the Socratic community can create.

Oct 14, 2017

#### Explanation:

A hyperbola is the set of all points in a plane such that the difference of the distances from two given fixed points called foci is constant.

Let the center of hyperbola be $\left(h , k\right)$ and focii along $y$-axis be $\left(h , k - c\right)$ and $\left(h , k + c\right)$. Let the difference of the distances be $2 a$.

Then from definition we should have

$\sqrt{{\left(x - h\right)}^{2} + {\left(y - k + c\right)}^{2}} - \sqrt{{\left(x - h\right)}^{2} + {\left(y - k - c\right)}^{2}} = 2 a$

or $\sqrt{{\left(x - h\right)}^{2} + {\left(y - k + c\right)}^{2}} = \sqrt{{\left(x - h\right)}^{2} + {\left(y - k - c\right)}^{2}} + 2 a$

and squaring both sides we get

${\left(x - h\right)}^{2} + {\left(y - k + c\right)}^{2} = {\left(x - h\right)}^{2} + {\left(y - k - c\right)}^{2} + 4 a \sqrt{{\left(x - h\right)}^{2} + {\left(y - k - c\right)}^{2}} + 4 {a}^{2}$

or ${y}^{2} + {k}^{2} + {c}^{2} + 2 y c - 2 y k - 2 k c = {y}^{2} + {k}^{2} + {c}^{2} - 2 y c - 2 y k + 2 k c + 4 a \sqrt{{\left(x - h\right)}^{2} + {\left(y - k - c\right)}^{2}} + 4 {a}^{2}$

or $4 y c - 4 k c = 4 a \sqrt{{\left(x - h\right)}^{2} + {\left(y - k - c\right)}^{2}} + 4 {a}^{2}$

or $a \sqrt{{\left(x - h\right)}^{2} + {\left(y - k - c\right)}^{2}} = y c - k c - {a}^{2}$

Squaring both sides again, we have

${a}^{2} \left({\left(x - h\right)}^{2} + {\left(y - k - c\right)}^{2}\right) = {\left(y c - k c - {a}^{2}\right)}^{2}$

or ${a}^{2} {\left(x - h\right)}^{2} + {a}^{2} {\left(y - k\right)}^{2} + {a}^{2} {c}^{2} \textcolor{red}{- 2 c y {a}^{2} + 2 k c {a}^{2}} = {y}^{2} {c}^{2} + {k}^{2} {c}^{2} + {a}^{4} \textcolor{red}{- 2 {a}^{2} y c + 2 {a}^{2} k c} - 2 y k {c}^{2}$

or ${a}^{2} {\left(x - h\right)}^{2} + {a}^{2} {\left(y - k\right)}^{2} = {y}^{2} {c}^{2} + {k}^{2} {c}^{2} + {a}^{4} - 2 y k {c}^{2}$

or ${a}^{2} {\left(x - h\right)}^{2} + {a}^{2} {\left(y - k\right)}^{2} = {c}^{2} \left({y}^{2} + {k}^{2} - 2 y k\right) + {a}^{4}$

or ${a}^{2} {\left(x - h\right)}^{2} + {a}^{2} {\left(y - k\right)}^{2} = {c}^{2} {\left(y - k\right)}^{2} + {a}^{4}$

or ${a}^{2} {\left(x - h\right)}^{2} - \left({c}^{2} - {a}^{2}\right) {\left(y - k\right)}^{2} = {a}^{4}$

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / \left({a}^{4} / \left({c}^{2} - {a}^{2}\right)\right) = 1$

or ${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / \left({b}^{2}\right) = 1$

where ${b}^{2} = {a}^{4} / \left({c}^{2} - {a}^{2}\right)$

• 10 minutes ago
• 14 minutes ago
• 23 minutes ago
• 24 minutes ago
• 14 seconds ago
• A minute ago
• 2 minutes ago
• 3 minutes ago
• 6 minutes ago
• 8 minutes ago
• 10 minutes ago
• 14 minutes ago
• 23 minutes ago
• 24 minutes ago