Question

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

(a) $\frac{1}{2}$

(b) $\sqrt{3}$

(c) $\frac{\sqrt{3}}{2}$

(d) $\frac{\sqrt{3}}{4}$

Answer 1

SOLUTION — Given equation can be rewritten as

$ 4(x-3)^2+16(y-1)^2=53 \\$

$\Rightarrow  \frac{(x-3)^2}{53 / 4}+\frac{(y-1)^2}{53 / 16}=1 \\$

$\text { Here, }  a^2=53 / 4, b^2=53 / 16 \\$

$\therefore  e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{4}{16}}=\frac{\sqrt{3}}{2}$

So, The correct option of this question will be (C).