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Question

Question: The focus of the parabola \({y^2} = 4y - 4x\) is a. (0,2) b. (1,2) c. (2,0) d. (2,1) ![](https...

The focus of the parabola y2=4y4x{y^2} = 4y - 4x is

a. (0,2)

b. (1,2)

c. (2,0)

d. (2,1)

Explanation

Solution

Hint: When we get these types of questions, firstly we’ll reduce the given equation to the standard form of that conic and then compare x0,y0{x_0}, {y_0} and aa with the standard equation of parabola. And then find the required parameter by putting values.

Complete step by step answer:

As, we know that the standard equation of parabola is (yy0)2=4a(xx0)(y - {y_0})^2 = 4a(x - {x_0}). In which,

\Rightarrow Vertex = (x0,y0)\left( {{x_0},{y_0}} \right) and,

\Rightarrow Focus of parabola is (x0+a,y0)\left( {{x_0} + a,{y_0}} \right)

Given Equation of parabola is y2=4y4x{y^2} = 4y - 4x

First we have to convert the given equation to the standard equation of parabola.

Taking 4y to LHS of the given equation it becomes,

y24y=4x\Rightarrow {y^2} - 4y = - 4x

Adding 4 both sides of the equation it becomes,

(y24y+4)=4x+4\Rightarrow \left( {{y^2} - 4y + 4} \right) = - 4x + 4

Taking - 4 common in RHS equation becomes,

\Rightarrow (y2)2=4(x1)({y - 2})^2 = - 4( x - 1 ) - (Eq 1)

Comparing equation 1with standard equation of parabola we get,

\Rightarrow x0=1,y0=2{x_0} = 1, {y_0} = 2 and a=1a = - 1

So, focus of the parabola in equation 1 will be,

\Rightarrow focus = (11,2)=(0,2)\left( {1 - 1,2} \right) = \left( {0,2} \right)

Hence the correct option for the question will be (a).

Note: Understand the diagram properly whenever you are facing these kinds of problems. A better knowledge of formulas will be an added advantage.