How do you find the vertices and foci of y^2/49-x^2=1?

Jan 26, 2017

Given the general Cartesian form:

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

The vertices are $\left(h , k - a\right) \mathmr{and} \left(h , k + a\right)$

The foci are $\left(h , k - \sqrt{{a}^{2} + {b}^{2}}\right) \mathmr{and} \left(h , k + \sqrt{{a}^{2} + {b}^{2}}\right)$

Explanation:

Write the given equation, ${y}^{2} / 49 - {x}^{2} = 1$, in the form of equation [1]:

Insert -0 with the square both numerators:

${\left(y - 0\right)}^{2} / 49 - {\left(x - 0\right)}^{2} = 1$

Write the 49 as ${7}^{2}$ and write a ${1}^{2}$ under the x term:

${\left(y - 0\right)}^{2} / {7}^{2} - {\left(x - 0\right)}^{2} / {1}^{2} = 1 \text{ [2]}$

Matching the variables in equation [1] with the values in equation [2], $h = 0 , k = 0 , a = 7 , \mathmr{and} b = 1$

Compute $\sqrt{{a}^{2} + {b}^{2}} = \sqrt{{7}^{2} + {1}^{2}} = \sqrt{50} = 5 \sqrt{2}$

Now, it is a simple matter to write the vertices and foci.

Vertices: $\left(0 , - 7\right) \mathmr{and} \left(0 , 7\right)$

Foci: $\left(0 , - 5 \sqrt{2}\right) \mathmr{and} \left(0 , 5 \sqrt{2}\right)$