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Question: Equation of the parabola with its vertex at \[\left( {1,1} \right)\] and focus \[\left( {3,1} \right...

Equation of the parabola with its vertex at (1,1)\left( {1,1} \right) and focus (3,1)\left( {3,1} \right) is
A. (x−1)2=8(y−1){(x - 1)^2} = 8(y - 1)
B. (y−1)2=8(x−3){(y - 1)^2} = 8(x - 3)
C. (y−1)2=8(x−1){(y - 1)^2} = 8(x - 1)
D. (x−3)2=8(y−1){(x - 3)^2} = 8(y - 1)

Explanation

Solution

First, we will find the directrix of the parabola for which focus (3,1)  \left( {3,1} \right)\; and vertex (1,1)  \left( {1,1} \right)\; are given, then we will use the general equation of a parabola when the directrix and focus are known to us to find the required equation. Hence, we will use (x−a)2+(y−b)2=directrix2{(x - a)^2} + {(y - b)^2} = directri{x^2} to find the equation. Then we will convert the equation of the parabola into its standard form and then we will compare the standard equation with the MCQs to get the final answer.

Complete step by step solution:
According to the question, we can imagine a parabola with the given parameters as

Here,
V→V \to vertex of the parabola
F→F \to focus of the parabola

And as the vertex is equidistant from foci and directrix, and latter is perpendicular to the axis of symmetry.

We get the equation of directrix as
⇒X=−1\Rightarrow X = - 1
Hence, the directrix is X+1=0X + 1 = 0
We know that equation of a parabola whose focus F(a,b)F\left( {a,b} \right) and directrix X+1=0X + 1 = 0 is know is given by
(x−a)2+(y−b)2=directrix2{(x - a)^2} + {(y - b)^2} = directri{x^2} … (1)

Parabola is the locus of a point whose distance from directrix x+1=0x + 1 = 0 and focus (3,1)(3,1) , Hence
We get a=3a = 3 , b=1b = 1 and directrix x=X+1x = X + 1 put values in (1), we get
(x−3)2+(y−1)2=(x+1)2{(x - 3)^2} + {(y - 1)^2} = {(x + 1)^2} … (2)

We know that,
(a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab
(a−b)2=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2ab

Hence, further solving (2), we get
⇒x2+9−6x+y2+1−2y=x2+1+2x\Rightarrow {x^2} + 9 - 6x + {y^2} + 1 - 2y = {x^2} + 1 + 2x

Cancelling out equal terms from, LHS and RHS, we get
⇒9−8x+y2−2y=0\Rightarrow 9 - 8x + {y^2} - 2y = 0
On rearranging we get,
⇒y2−2y+9=8x\Rightarrow {y^2} - 2y + 9 = 8x

Completing the square in LHS, we get
⇒y2−2y+1+8=8x\Rightarrow {y^2} - 2y + 1 + 8 = 8x
Using (a−b)2=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2ab , we get,
⇒(y−1)2=8x−8\Rightarrow {(y - 1)^2} = 8x - 8
Taking 8 common on RHS we get,
⇒(y−1)2=8(x−1)\Rightarrow {(y - 1)^2} = 8(x - 1)

Hence, the required equation is (y−1)2=8(x−1){(y - 1)^2} = 8(x - 1)
Therefore, the final answer is option C.

Note:
These types of questions can be easily solved if we can visually illustrate the required curve in a 2D plane because after that many hidden information is gained, for example in this particular question, the equation of directrix was found as a key to solve the given question easily.