Question

Latus rectum of ellipse $4{x^2} + 9{y^2} - 8x - 36y + 4 = 0$
$
  {\text{A}}{\text{.}}\dfrac{8}{3} \\
  {\text{B}}{\text{.}}\dfrac{4}{3} \\
  {\text{C}}{\text{.}}\dfrac{{\sqrt 5 }}{3} \\
  {\text{D}}{\text{.}}\dfrac{{16}}{3} \\
$

Answer 1

Hint: In this question compare the given equation representing the ellipse with the standard equation of the ellipse and then find the values of a and b for getting the latus rectum.

Complete step-by-step answer:
Given equation:
$
  4{x^2} + 9{y^2} - 8x - 36y + 4 = 0 $
  $\Rightarrow 4{x^2} - 8x + 9{y^2} - 36y + 4 = 0$
taking 4 common from \[4{x^2} - 8x\] and 9 common from $9{y^2} - 36y$ then equation can be written as:
$4\left( {{x^2} - 2x + 1} \right) - 4 + 9\left( {{y^2} - 4y + 4} \right) - 36 + 4 = 0$
now $\left( {{x^2} - 2x + 1} \right)$ can be written as ${\left( {x - 1} \right)^2}$
and $\left( {{y^2} - 4y + 4} \right)$ can be written as ${\left( {y - 2} \right)^2}$
therefore complete equation can now be written as:
$4{\left( {x - 1} \right)^2} + 9{\left( {y - 2} \right)^2} = 36$
divide both side by 36 then we get,
$\dfrac{{{{\left( {x - 1} \right)}^2}}}{9} + \dfrac{{{{\left( {y - 2} \right)}^2}}}{4} = 1$
now compare with standard equation where ${\text{h,k}}$ are the coordinate of the centre of ellipse
$
  \dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 \\
  {a^2} = 9 \Rightarrow a = 3 \\
  {b^2} = 4 \Rightarrow b = 2 \\
$
Formula for Latus rectum $ = \dfrac{{2{b^2}}}{a}$
putting values of $a = 2$ and $b = 3$
$ \Rightarrow \dfrac{{2{b^2}}}{a} = \dfrac{{2 \times 4}}{3} = \dfrac{8}{3}$
Hence the required value is $\dfrac{8}{3}$.

Note: In this question first we simplified the given equation of ellipse in a way that it can be compared to the standard equation of ellipse and hence we found the value of $a$ and $b$, after that we put the value of these two component in the formula of latus rectum and got the required value.