Question

If Y-axis is the directrix of the ellipse with eccentricity $e={\displaystyle \frac{1}{2}}$ and the corresponding focus is at (3,0), find the equation to its auxiliary circle.
$$\begin{gathered} \text{A}\text{. }{x}^2+y^2-8x+12=0 \\ \text{B}\text{. }{x}^2+y^2-8x-12=0 \\ \text{C}\text{. }{x}^2+y^2-8x+9=0 \\ \text{D}\text{. }{x}^2+y^2=4 \end{gathered}$$

Answer 1

Hint- Here, we will proceed by using the formulas for the equation of directrix i.e.,$x=h-{\displaystyle \frac{a}{e}}$ , the focus coordinates i.e., F(h-ae,k) and $b^2=a^2\left(1-e^2\right)$ corresponding to any ellipse ${\displaystyle \frac{{\left(x-h\right)}^2}{a^2}}+{\displaystyle \frac{{\left(y-k\right)}^2}{b^2}}=1$ where a>b.

Complete step by step answer:

Given, Directrix to the given ellipse is represented by equation of Y-axis i.e., x = 0
Eccentricity, $e={\displaystyle \frac{1}{2}}$
Focus of the ellipse is at F(3,0)
Since, the directrix of the given ellipse lies along the Y-axis. Therefore, the ellipse will be oriented along the X-axis.
Let the equation of the ellipse along Y-axis is given by ${\displaystyle \frac{{\left(x-h\right)}^2}{a^2}}+{\displaystyle \frac{{\left(y-k\right)}^2}{b^2}}=1\to \text{(1)}$ where a>b and the centre of the ellipse lies at point C(h,k).
As we know that the equation of the directrix to any ellipse ${\displaystyle \frac{{\left(x-h\right)}^2}{a^2}}+{\displaystyle \frac{{\left(y-k\right)}^2}{b^2}}=1$ (having eccentricity as e) where a>b is given by
$x=h-{\displaystyle \frac{a}{e}}$
For the given ellipse, equation of the directrix is x = 0
$$\begin{gathered} \Rightarrow 0=h-{\displaystyle \frac{a}{e}} \\ \Rightarrow h={\displaystyle \frac{a}{e}} \end{gathered}$$
By putting $e={\displaystyle \frac{1}{2}}$ in the above equation, we get
$$\begin{gathered} \Rightarrow h={\displaystyle \frac{a}{\left({\displaystyle \frac{1}{2}}\right)}} \\ \Rightarrow h=2a\to \text{(2)} \end{gathered}$$
Also, the focus coordinates for any ellipse ${\displaystyle \frac{{\left(x-h\right)}^2}{a^2}}+{\displaystyle \frac{{\left(y-k\right)}^2}{b^2}}=1$ (having eccentricity as e) where a>b is given by F(h-ae,k)
Also, focus of the given ellipse is F(3,0)
So, h-ae = 3
$$\begin{gathered} \Rightarrow 2a-a\left({\displaystyle \frac{1}{2}}\right)=3 \\ \Rightarrow 2a-{\displaystyle \frac{a}{2}}=3 \\ \Rightarrow {\displaystyle \frac{4a-a}{2}}=3 \\ \Rightarrow {\displaystyle \frac{3a}{2}}=3 \\ \Rightarrow a=2 \end{gathered}$$
Hence, $a^2=2^2=4$
Also, k = 0
Using the formula $b^2=a^2\left(1-e^2\right)$, the value of $b^2$ is given as
$$\begin{gathered} \Rightarrow b^2=4\left(1-{\left({\displaystyle \frac{1}{2}}\right)}^2\right)=4\left(1-{\displaystyle \frac{1}{4}}\right)=4\left({\displaystyle \frac{4-1}{4}}\right)=4\left({\displaystyle \frac{3}{4}}\right) \\ \Rightarrow b^2=3 \end{gathered}$$
Putting a = 2 in equation (2), we get
$\Rightarrow h=2\times 2=4$
Putting $a^2=4$, $b^2=3$, h = 4 and k = 0 in equation (1), we get
$$\begin{gathered} \Rightarrow {\displaystyle \frac{{\left(x-4\right)}^2}{4}}+{\displaystyle \frac{{\left(y-0\right)}^2}{3}}=1 \\ \Rightarrow {\displaystyle \frac{{\left(x-4\right)}^2}{4}}+{\displaystyle \frac{y^2}{3}}=1 \end{gathered}$$
This above equation represents the equation of the given ellipse.
Equation of the auxiliary circle to the ellipse ${\displaystyle \frac{{\left(x-h\right)}^2}{a^2}}+{\displaystyle \frac{{\left(y-k\right)}^2}{b^2}}=1$ where a>b is given by
${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2\to \text{(3)}$
By putting $a^2=4$, h = 4 and k = 0 in the equation (3), we get
$$\begin{gathered} \Rightarrow {\left(x-4\right)}^2+{\left(y-0\right)}^2=4 \\ \Rightarrow {\left(x-4\right)}^2+y^2=4 \\ \Rightarrow x^2+4^2-8x+y^2=4 \\ \Rightarrow x^2+y^2-8x+12=0 \end{gathered}$$
The above equation represents the required equation of the auxiliary circle to the given ellipse.
Hence, option A is correct.

Note- In this particular problem, firstly it very important to find out that the given ellipse corresponds to which one of the two general cases of ellipse i.e., ${\displaystyle \frac{{\left(x-h\right)}^2}{a^2}}+{\displaystyle \frac{{\left(y-k\right)}^2}{b^2}}=1$ where a>b or ${\displaystyle \frac{{\left(x-h\right)}^2}{a^2}}+{\displaystyle \frac{{\left(y-k\right)}^2}{b^2}}=1$ where b>a, in order to use the formulas for various parameters. Also, the major axis of the given ellipse is X-axis and the minor axis is Y-axis.