Question

How do you solve the standard form of an ellipse given foci: $\left( {0, - 3} \right),\left( {0,3} \right)$ vertices: $\left( {0, - 4} \right),\left( {0,4} \right)$ ?

Answer 1

Hint : As you can see the foci and the vertices both lie on the vertical axis , the ellipse is having its major axis as the vertical axis . The standard form of ellipse having major axis as vertical axis is $\dfrac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} + \dfrac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} = 1$ . Determine the centre $C\left( {h,k} \right)$ and value for variables $a,b$ using the properties of ellipse.

Complete step by step solution:
We are given foci $\left( {0, - 3} \right),\left( {0,3} \right)$ and vertices $\left( {0, - 4} \right),\left( {0,4} \right)$ for some ellipse.
As per the question, we have to obtain an equation in standard form for a ellipse having foci at coordinates $\left( {0, - 3} \right),\left( {0,3} \right)$ and vertices at $\left( {0, - 4} \right),\left( {0,4} \right)$ .
Since, the foci and vertices are on the y-axis of the plane, which shows that the major axis of the ellipse is on the y-axis. So the standard equation form for ellipse having major axis on y-axis is $\dfrac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} + \dfrac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} = 1$ where $\left( {h,k} \right)$ is the centre of the ellipse.
Now let's find out the values for variables $a,b,h,k$ .
Distance ${D_f}$ between the foci for ellipse $2c$
Using the distance formula $D = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}}$ to find the distance between foci $\left( {0, - 3} \right)\,and\,\left( {0,3} \right)$ , we obtain ${D_f}$ as
${D_f} = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} \\\ {D_f} = \sqrt {{{\left( {0 - 0} \right)}^2} + {{\left( { - 3 - 3} \right)}^2}} \\\ {D_f} = \sqrt {{{\left( { - 6} \right)}^2}} \\\ {D_f} = 6 \\\ \Rightarrow 2c = 6 \Rightarrow c = 3 \;$
Now Similarly we know that the distance ${D_v}$ between the vertices for ellipse is $2a$
Again using the distance formula for distance between foci point $\left( {0, - 4} \right)\,and\,\left( {0,4} \right)$ and finding the value for variable $a$ as
${D_v} = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} \\\ {D_v} = \sqrt {{{\left( {0 - 0} \right)}^2} + {{\left( { - 4 - 4} \right)}^2}} \\\ {D_v} = \sqrt {{{\left( { - 8} \right)}^2}} \\\ {D_v} = 8 \\\ \Rightarrow 2a = 8 \Rightarrow a = 4 \;$
Now using the values for $a = 4\,and\,c = 3$ , find the value for variable $b$ by using the relation ${c^2} = {a^2} - {b^2}$ , we get

$${c^2} = {a^2} - {b^2} \\\ \Rightarrow {b^2} = {a^2} - {c^2} \\\ \Rightarrow {b^2} = {4^2} - {3^2} \\\ \Rightarrow {b^2} = 16 - 9 \\\ \Rightarrow {b^2} = 7 \\\ \Rightarrow b = \sqrt 7 \;$$

Thus, we have obtained the value for $a,b,c$ as $4,\sqrt 7 ,3$ respectively.
Now let’s find out the centre $\left( {h,k} \right)$ of the ellipse using the relation that
Midpoint $M$ between the foci or vertices $=$ centre $C\left( {h,k} \right)$ of the ellipse
Using the coordinates of foci we have
$M:\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right) = C\left( {h,k} \right) \\\ M:\left( {\dfrac{0}{2},\dfrac{{ - 3 + 3}}{2}} \right) = C\left( {h,k} \right) \\\ M:\left( {0,0} \right) = C\left( {h,k} \right) \;$
Hence, we have the centre of the ellipse $C\left( {h,k} \right)$ equal to the origin $\left( {0,0} \right)$ .
Now Putting all the values of variables into the standard form of the ellipse specified in the equation (1), we have
$\dfrac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} + \dfrac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} = 1 \\\ \dfrac{{{{\left( {x - 0} \right)}^2}}}{{{{\left( {\sqrt 7 } \right)}^2}}} + \dfrac{{{{\left( {y - 0} \right)}^2}}}{{{{\left( 4 \right)}^2}}} = 1 \;$
Simplifying further ,we get
$\Rightarrow \dfrac{{{x^2}}}{7} + \dfrac{{{y^2}}}{{16}} = 1$
Therefore, standard equation of the ellipse having foci $\left( {0, - 3} \right),\left( {0,3} \right)$ and vertices $\left( {0, - 4} \right),\left( {0,4} \right)$ is equal to $\dfrac{{{x^2}}}{7} + \dfrac{{{y^2}}}{{16}} = 1$
Formula:
$M:\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
$D = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}}$

Note : 1.. Every ellipse has two axes of symmetry. The longer axis is called the major axis and the shorter axis is called the minor axis.
2. Each end point of the major axis is the vertex of the ellipse and the end point on the minor axis are called the co-vertices .
3. Remember that the foci for the ellipse always lie on the major axis and not on the minor axis.
4. You can also use the vertices of the ellipse to find out the centre . The result will be the same whether you are taking vertices or foci coordinates.