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Find the vertex of the parabola \[{x^2} + 2y = 8x - 7\] .
A. \[\left( {\dfrac{9}{2},0} \right)\]
B. \[\left( {4,\dfrac{9}{2}} \right)\]
C. \[\left( {2,\dfrac{9}{2}} \right)\]
D. \[\left( {4,\dfrac{7}{2}} \right)\]

Answer
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161.1k+ views
Hint: Rewrite the given equation in the general form of a parabola and obtain the required vertex. Subtract 8x from both sides of the given equation, then add 16 to both sides of the obtained equation. Factor out the term -2 from the expression \[2y - 9\] to obtain the required solution.

Formula used:
The general form of a parabola is,
\[{\left( {x - h} \right)^2} = 4a(y - k)\] , where \[(h,k)\] is the vertex.

Complete step to step solution:
The given equation is,
\[{x^2} + 2y = 8x - 7\]
Subtract 8x from both sides of the equation \[{x^2} + 2y = 8x - 7\],
\[{x^2} - 8x + 2y = - 7\]
Subtract 2y from both sides of the equation \[{x^2} - 8x + 2y = - 7\],
\[{x^2} - 8x = - 7 - 2y\]
\[{x^2} - 8x + 16 = - 7 - 2y + 16\]
\[{x^2} - 2.4.x + {4^2} = 9 - 2y\]
\[{(x - 4)^2} = - 2(y - \dfrac{9}{2})\]
Compare the equation \[{(x - 4)^2} = - 2(y - \dfrac{9}{2})\]with \[{\left( {x - h} \right)^2} = 4a(y - k)\] to obtain the vertex.
\[h = 4\] and \[k = \dfrac{9}{2}\]
Therefore, the vertex is \[\left( {4,\dfrac{9}{2}} \right)\].
The correct option is B.

Additional information :
The intersection point of parabola and symmetric line is the vertex of a parabola.
For a vertical parabola, the vertex is the lowest or the highest point of the parabola.

Note: To solve this type of question, we have to identify whether the parabola is a horizontal parabola or vertical parabola. The highest power of x is 2. So, the given parabola is a horizontal parabola. Now rewrite the equation in the form \[{\left( {x - h} \right)^2} = 4a(y - k)\] and compare to find the vertex of the parabola