Question

The eccentricity of the conic $4x^2+16y^2-24x-32y=1$ is:

• A. $\frac{1}{2}$
• B. $\sqrt{3}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{\sqrt{3}}{4}$

Answer 1

The correct option is C

$\frac{\sqrt{3}}{2}$

Explanation for the correct option:

Finding the eccentricity:

The given equation of conic is

$4x^2+16y^2-24x-32y=1$

$$\begin{gathered} \left(4x^2-24x+36-36\right)+\left(16y^2-32y+16-16\right)=1 \\ \left[4\right(x^2–6x+9)–36]+[16\left([y^2–2y+1\right)–16]=1 \\ 4{(x–3)}^2+16{(y–1)}^2=53 \\ \frac{{[x–3]}^2}{{\displaystyle \frac{53}{4}}}+\frac{{[y–1]}^2}{{\displaystyle \frac{53}{16}}}=1 \end{gathered}$$

The above equation of the ellipse is of the form

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ having eccentricity $e=\sqrt{1-\frac{b^2}{a^2}}$

After comparing the above equations we have,

$a^2=\frac{53}{2},b^2=\frac{53}{16}$

Therefore,

$$\begin{gathered} \Rightarrow e=\sqrt{1-\frac{{\displaystyle \frac{53}{16}}}{{\displaystyle \frac{53}{4}}}} \\ =\sqrt{1-\frac{1}{4}} \\ =\sqrt{\frac{3}{4}} \\ =\frac{\sqrt{3}}{2} \end{gathered}$$

Hence, option (C) is the correct answer.