Question

The length of latus rectum of the parabola \[4{{y}^{2}}+3x+3y+1=0\] is
(a) \[\dfrac{-3}{4}\]
(b) 7
(c) 12
(d) \[\dfrac{3}{4}\]

Answer 1

Hint: The general equation of parabola is \[{{y}^{2}}=4ax\]. Formulate the given equation of parabola into the form of \[{{y}^{2}}=4ax\]. 4a represents the latus rectum.

Complete step-by-step answer:
We have been given the equation of parabola,

\[4{{y}^{2}}+3x+3y+1=0\]

We know the general equation of parabola, \[{{y}^{2}}=4ax\].

In this expression, 4a represents the latus rectum.

Now consider the given equation,

\[\begin{align}

  & 4{{y}^{2}}+3x+3y+1=0 \\

 & \Rightarrow 4{{y}^{2}}+3y=-3x-1 \\

\end{align}\]

Let’s divide the equation by 4.

\[\therefore {{y}^{2}}+\dfrac{3}{4}y=-\left[ \dfrac{3}{4}x+\dfrac{1}{4} \right]\]

Add \[\dfrac{9}{64}\] on both sides of the equation.

\[{{y}^{2}}+\dfrac{3}{4}y+\dfrac{9}{64}=-\left[ \dfrac{3}{4}x+\dfrac{1}{4} \right]+\dfrac{9}{64}\]
We add \[\dfrac{9}{64}\] to convert LHS into \[{{\left( a+b \right)}^{2}}\] form.

\[\begin{align}
  & {{y}^{2}}+\dfrac{3}{4}y+{{\left( \dfrac{3}{8} \right)}^{2}}=\dfrac{-3}{4}x-\dfrac{1}{4}+\dfrac{9}{64} \\
 & {{\left( y+\dfrac{3}{8} \right)}^{2}}=\left[ \dfrac{-3}{4}x-\dfrac{7}{64} \right] \\
 & {{\left( y+\dfrac{3}{8} \right)}^{2}}=\dfrac{3}{4}\left[ -x-\dfrac{7}{64} \right] \\
 & {{\left( y+\dfrac{3}{8} \right)}^{2}}=\dfrac{-3}{4}\left( x+\dfrac{7}{64} \right) \\
\end{align}\]

Now, this above equation is of the form, \[{{y}^{2}}=4ax\].

We know that the length of the latus rectum \[=4a=\dfrac{3}{4}\].

Thus we got the latus rectum as \[\dfrac{3}{4}\] for the given equation.

Thus option (d) is the correct answer.

Note: Remember that to find the latus rectum, you need to formulate the given equation in the form of the general equation of parabola. Be careful with signs and don’t mix them up and you will get the answer wrong. When you get a latus rectum the whole value of 4a is \[\dfrac{3}{4}\]. Don’t try to simplify and find a.