Question

$A{y^2} + By + Cx + D = 0,\left( {ABC \ne 0} \right)$ be the equation of a parabola. Which of the options is correct?
(A) Then the length of the latus rectum is $\left| {\dfrac{C}{A}} \right|$
(B) The axis of the parabola is vertical
(C) The y-coordinate of the vertex is $- \dfrac{B}{{2A}}$
(D) The x-coordinate of the vertex is $\dfrac{D}{A} + \dfrac{{{B^2}}}{{4AC}}$

Answer 1

Start with transforming the given equation of the parabola into its general form, i.e. ${Y^2} = \pm 4AX$. Complete the square of the y-coordinate using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$. Now compare the newly formed equation with its general form. Check for the characteristics discussed in the options and find the answer.

Complete step-by-step answer:
We are given with an equation of a parabola $A{y^2} + By + Cx + D = 0,\left( {ABC \ne 0} \right)$and we have to find the length of latus rectum, its y-coordinate, its x-coordinate and position of an axis according to the options given.
So, let’s just begin with changing this equation into a general form of a parabola, i.e. ${Y^2} = \pm 4aX{\text{ or }}{{\text{X}}^2} = \pm 4aY$
$\Rightarrow A{y^2} + By + Cx + D = 0 \Rightarrow {y^2} + \dfrac{B}{A}y + \dfrac{C}{A}x + \dfrac{D}{A} = 0$
We will try to make a perfect square of y to make in the form ${Y^2} = \pm 4aX$
$\Rightarrow {y^2} + 2 \times \dfrac{1}{2} \times \dfrac{B}{A}y + \dfrac{C}{A}x + \dfrac{D}{A} = 0 \Rightarrow \left( {{y^2} + 2 \times \dfrac{B}{{2A}}y + \dfrac{{{B^2}}}{{4A}}} \right) - \dfrac{{{B^2}}}{{4A}} + \dfrac{C}{A}x + \dfrac{D}{A} = 0$
Now, we use the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ in the above transformed equation as:
$\Rightarrow {\left( {y + \dfrac{B}{{2A}}} \right)^2} = - \dfrac{{Cx}}{A} - \dfrac{D}{A} + \dfrac{{{B^2}}}{{4{A^2}}} = - \dfrac{C}{A}\left( {x + \dfrac{D}{C} - \dfrac{{{B^2}}}{{4AC}}} \right)$
Therefore, we got the final equation of parabola as $\Rightarrow {\left( {y + \dfrac{B}{{2A}}} \right)^2} = - \dfrac{C}{A}\left( {x + \dfrac{D}{C} - \dfrac{{{B^2}}}{{4AC}}} \right)$ (1)
But we know that a general equation of a parabola is of the form ${Y^2} = \pm 4aX$ (2)
And in general form the length of the latus rectum of the parabola$= 4a$
So, by comparing equation (1) and (2), we get: $\Rightarrow 4a = - \dfrac{C}{A}$
Therefore, the length of the latus rectum is $\left| {\dfrac{C}{A}} \right|$
For a parabola of form${Y^2} = \pm 4aX$, the x-axis will be the axis of symmetry. Therefore, we can say that the axis of this parabola is horizontal.
For finding x-coordinate and y-coordinate, we just need to put $X = 0{\text{ and }}Y = 0$
From comparing (1) and (2) again, we get: $X = 0 \Rightarrow x + \dfrac{D}{C} - \dfrac{{{B^2}}}{{4AC}} = 0 \Rightarrow x = - \dfrac{D}{C} + \dfrac{{{B^2}}}{{4AC}}$ and similarly for y: $Y = 0 \Rightarrow y + \dfrac{B}{{2A}} = 0 \Rightarrow y = - \dfrac{B}{{2A}}$
Hence, we can say from the above results that the option (A) and (C) are correct.

So, the correct answers are “Option A” and “Option C”.

Note: Try to be careful when transforming the given equation into the general form. Notice the transformations of multiplication and division of $\dfrac{B}{A}y$ by $2$ , and the addition and subtraction of $\dfrac{{{B^2}}}{{4A}}$ in the equation in order to make a perfect square of y-coordinate. Completing the square using ${a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}$was a crucial part of the solution.