Question

If the equation ${\left( {2x - 3 - y} \right)^2} = - 20\left( {x + 2y - 4} \right)$ represents a parabola then its vertex is
A. $\left( {3, - 3} \right)$
B. $\left( {6, - 1} \right)$
C. $\left( {2,1} \right)$
D. $\left( {4,4} \right)$

Answer 1

Hint: To solve such questions start by writing the standard form of a parabola. After that compare the given equation with the standard form of the parabola. Finally, solve the equations obtained to find the vertex of the given parabola.

Complete step-by-step solution:
Given that the equation ${\left( {2x - 3 - y} \right)^2} = - 20\left( {x + 2y - 4} \right)$ represents a parabola.
We know that the standard form of a parabola is ${Y^2} = 4aX$
Next, compare this equation to the given equation. That is,
$X = x + 2y - 4$
$Y = 2x - y - 3$
$a = - 5$
It can be seen that the line $X = 0$ and $Y = 0$ are perpendicular to each other.
It is also known that the vertex of a parabola ${Y^2} = 4aX$is $X = 0$and $Y = 0$ .
Therefore, it can be said that the vertex of the parabola will be the solution of the equations $x + 2y - 4 = 0$ and $2x - y - 3 = 0$ .
So the next step is to solve these equations. That is,
Consider the equation $x + 2y - 4 = 0$ .
Rearranging this we get,
$x = - 2y + 4......(1)$
Substituting equation $(1)$ in the equation $2x - y - 3 = 0$ . That is,
$2\left( { - 2y + 4} \right) - y - 3 = 0$
$- 4y + 8 - y - 3 = 0$
Simplifying further we get,
$- 5y = - 5$
Dividing both the LHS and RHS by $5$ and canceling out the minus sign on both sides, we get,
$y = 1$
Next substitute the value $y = 1$ in the equation $(1)$ , we get,
$x = - 2\left( 1 \right) + 4$
Further simplifying we get,
$x = - 2 + 4$
$x = 2$
Therefore, the vertex of the given parabola is $\left( {2,1} \right)$ .

Hence option C is correct.

Note: The fixed point on the parabola is known as the locus. The fixed-line on the parabola is known as the directrix. The chord which passes through the focus and is perpendicular to the axis of the parabola is known as the latus rectum.