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Question: Find the equation of the parabola having the vertex at \( \left( {0,1} \right) \) and the focus at \...

Find the equation of the parabola having the vertex at (0,1)\left( {0,1} \right) and the focus at (0,0)\left( {0,0} \right) :
A. x2+4y4=0{x^2} + 4y - 4 = 0
B. x2+4y+4=0{x^2} + 4y + 4 = 0
C. x24y+4=0{x^2} - 4y + 4 = 0
D. x24y4=0{x^2} - 4y - 4 = 0

Explanation

Solution

Hint : Take the general equation of the parabola and then substitute the values of the vertex. Calculate the value of aa by taking the distance of the vertex from the focus and then substitute in the equation of parabola.

Complete step-by-step answer :
As given in the question, the parabola has the vertex at (0,1)\left( {0,1} \right) and the focus at (0,0)\left( {0,0} \right) . The focus lies below the vertex. So, the general equation of the parabola is (xh)2=4a(yk){\left( {x - h} \right)^2} = - 4a\left( {y - k} \right) where (h,k)\left( {h,k} \right) is the vertex of the parabola and aa is the distance between the vertex and the focus.
Substitute 00 for hh and 11 for kk in the equation of parabola as per given in the question:
(xh)2=4a(yk) (x0)2=4a(y1) x2=4a(y1)              (1)   {\left( {x - h} \right)^2} = - 4a\left( {y - k} \right) \\\ {\left( {x - 0} \right)^2} = - 4a\left( {y - 1} \right) \\\ {x^2} = - 4a\left( {y - 1} \right)\;\;\;\;\;\;\; \ldots \left( 1 \right) \;
Now calculate the distance between the vertex (0,1)\left( {0,1} \right) and focus (0,0)\left( {0,0} \right) by using the distance formula.
(00)2+(01)2=02+(1)2 =1 =1   \sqrt {{{\left( {0 - 0} \right)}^2} + {{\left( {0 - 1} \right)}^2}} = \sqrt {{0^2} + {{\left( { - 1} \right)}^2}} \\\ = \sqrt 1 \\\ = 1 \;
As the value of aa is the distance of the focus from the vertex which is equal to 11 .
Substitute 11 for aa in the equation (1)\left( 1 \right) of parabola.
x2=4a(y1) x2=4(1)(y1) x2=4y+4 x2+4y4=0   {x^2} = - 4a\left( {y - 1} \right) \\\ {x^2} = - 4\left( 1 \right)\left( {y - 1} \right) \\\ {x^2} = - 4y + 4 \\\ {x^2} + 4y - 4 = 0 \;
So, the equation of the parabola is equal to x2+4y4=0{x^2} + 4y - 4 = 0 .
So, the correct answer is “Option A”.

Note : The general equation of the parabola with focus below the vertex is equal to (xh)2=4a(yk){\left( {x - h} \right)^2} = - 4a\left( {y - k} \right) where (h,k)\left( {h,k} \right) is the vertex of the parabola and aa is the distance between the vertex and the focus. The distance between two points (x1,y1)\left( {{x_1},{y_1}} \right) and (x2,y2)\left( {{x_2},{y_2}} \right) is equal to (x2x1)2+(y2y1)2\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} with the help of the distance formula in two dimensional geometry.