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# Find equation of latus rectum of the parabola ${\left( {x + 1} \right)^2} = 32y$.A. $y = 32$B. $x - 8 = 0$C. $y - 8 = 0$D. $x = 32$

Last updated date: 04th Aug 2024
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Hint: Here we will first find the value of $a$ by comparing the given parabola equation with the standard equation of parabola. Then we will use this to get the value of the focus point of the parabola. Then we will use the concept of the latus rectum to get the equation of the latus rectum of the given parabola.

Complete step by step solution:
Then the given equation of parabola is ${\left( {x + 1} \right)^2} = 32y$.
First, we will compare the equation with the standard form of parabola i.e. ${x^2} = 4ay$ to find the value of $a$. Therefore by comparing we get
$4a = 32$
Dividing both sides by 4, we get
$\Rightarrow a = \dfrac{{32}}{4} = 8$
Now we will find the focus of the given parabola.
We know that the focus point of the parabola is $\left( { - 1,a} \right)$.
Now by putting the value of $a$, we can write that the focus point of the parabola is $\left( { - 1,8} \right)$.
We know that the latus rectum of the parabola passes through the focus point of the parabola and also perpendicular to the axis of parabola. Therefore, by using this we get
The equation of the latus rectum of the parabola is $y - 8 = 0$.

So, option C is the correct option.

Note:
Here we should note that if the equation of the parabola is ${y^2} = 4ax$, then the parabola is according to the x-axis and is the equation of the parabola is ${x^2} = 4ay$, then the parabola is according to the y-axis. Parabola is a set of points in the Cartesian plane whose distance from a point (focus of the parabola) is equal to the distance from a particular line and this line is generally known as the directrix of the parabola. Directrix of the parabola is always perpendicular to the line of symmetry of the parabola. We should know that the equation of the normal of the parabola should satisfy any point which lies on the curve of parabola.