Question

How do you write an equation of an ellipse with centre $\left( 0,4 \right)$ and a=2c, vertices $\left( -4,4 \right)\left( 4,4 \right)$.

Answer 1

In this problem, we have to find the equation of an ellipse with the given centre and the vertices. We know that the standard form of the ellipse is $\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$ , a and b are the semi major and the minor axis. We know that the relation to determine the value of a and b is ${{c}^{2}}={{a}^{2}}-{{b}^{2}}$ . we can now find the value of a and b and we have a centre, by substituting those values, we can find the equation of the given ellipse.

Complete step by step answer:
We know that the standard form of the ellipse is,
$\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1,a>b$…… (1)
Where, $\left( h,k \right)$ is the centre and a, b are the semi major and the semi minor axis respectively.
We are given that the centre is,
$\left( h,k \right)=\left( 0,4 \right)$……. (2)
Now we can find the value of a and b.
We know that for the equation of ellipse, a is strictly greater than b, for this semi major and semi minor axis, we have a relation through which we can determine them, it sis
${{c}^{2}}={{a}^{2}}-{{b}^{2}}$
We are given that, a = 2c. we can now substitute this in the above relation, we get

& \Rightarrow {{c}^{2}}={{\left( 2c \right)}^{2}}+{{b}^{2}} \\\ & \Rightarrow {{c}^{2}}=4{{c}^{2}}+{{b}^{2}} \\\ & \Rightarrow {{b}^{2}}=3{{c}^{2}}.....(3) \\\ \end{aligned}$$ We know that the given vertices are $$\left( -4,4 \right)\left( 4,4 \right)$$. Now from looking at the vertices and the centre, we can see that the major semi axis length, a is 4. We can now substitute a = 4 in a = 2c, we get $$\Rightarrow c=2$$ Now we can substitute the above c value in (3), we get $$\begin{aligned} & \Rightarrow {{b}^{2}}=3{{\left( 2 \right)}^{2}} \\\ & \Rightarrow b=\sqrt{12}=2\sqrt{3} \\\ \end{aligned}$$ We can now substitute the centre, $$\left( h,k \right)=\left( 0,4 \right)$$ , a= 4 and b = $$2\sqrt{3}$$ in (1), we get $$\Rightarrow \dfrac{{{\left( x-0 \right)}^{2}}}{{{4}^{2}}}+\dfrac{{{\left( y-4 \right)}^{2}}}{{{\left( 2\sqrt{3} \right)}^{2}}}=1$$ We can now simplify the above step, we get $$\Rightarrow \dfrac{{{x}^{2}}}{16}+\dfrac{{{\left( y-4 \right)}^{2}}}{4}=1$$ Therefore, the required equation of ellipse is $$\dfrac{{{x}^{2}}}{16}+\dfrac{{{\left( y-4 \right)}^{2}}}{4}=1$$.  **Note:** We should know that the relation to determine the value of a and b is $${{c}^{2}}={{a}^{2}}-{{b}^{2}}$$ . We know that the standard form of the ellipse is, $$\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1,a>b$$ Where, $$\left( h,k \right)$$ is the centre and a, b are the semi major and the semi minor axis respectively.