Answer
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Hint:
Here, we are required to find the vertex of the given parabola. Thus, we will simplify the given equation and solve it to find its vertex form. Then, comparing it with the general equation of parabola, we will be able to find the required vertex of the parabola.
Formula Used:
\[{a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}\]
Complete step by step solution:
The given equation is \[{y^2} + 2y - 2x + 5 = 0\]
Now, we will shift all the terms to the RHS except the term containing the variable \[x\]
Hence, we get,
\[2x = {y^2} + 2y + 5\]
Dividing 2 on both sides, we get
\[ \Rightarrow x = \dfrac{{{y^2} + 2y + 5}}{2}\]
Rewriting the equation, we get
\[ \Rightarrow x = \dfrac{{{y^2} + 2y + 1}}{2} + \dfrac{4}{2}\]
Hence, using the identity \[{a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}\] in the numerator of first fraction in the RHS, we get,
\[ \Rightarrow x = \dfrac{{{{\left( {y + 1} \right)}^2}}}{2} + 2\]
Or \[x = \dfrac{1}{2}{\left( {y - \left( { - 1} \right)} \right)^2} + 2\]
Comparing this with \[x = m{\left( {y - b} \right)^2} + a\], where \[\left( {x,y} \right) = \left( {a,b} \right)\] is the vertex.
Thus, comparing the given equation with the general equation, we get,
Vertex \[\left( {x,y} \right) = \left( {2, - 1} \right)\]
Therefore, the vertex of the given parabola \[{y^2} + 2y - 2x + 5 = 0\] is \[\left( {2, - 1} \right)\]
Hence, this is the required answer.
Note:
A parabola is a curve having a focus and a directrix, such that each point on parabola is at equal distance from them. The axis of symmetry of a parabola is a line about which the parabola is symmetrical. When the parabola is vertical, the line of symmetry is vertical. When a quadratic function is graphed in the coordinate plane, the resulting parabola and corresponding axis of symmetry are vertical. Also, in this question, the parabola opens "sideways" and the axis of symmetry of the parabola is horizontal. The standard form of equation of a horizontal parabola is \[x = a{y^2} + by + c\] where \[a,b\] and \[c\] are all real numbers \[a \ne 0\].
Here, we are required to find the vertex of the given parabola. Thus, we will simplify the given equation and solve it to find its vertex form. Then, comparing it with the general equation of parabola, we will be able to find the required vertex of the parabola.
Formula Used:
\[{a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}\]
Complete step by step solution:
The given equation is \[{y^2} + 2y - 2x + 5 = 0\]
Now, we will shift all the terms to the RHS except the term containing the variable \[x\]
Hence, we get,
\[2x = {y^2} + 2y + 5\]
Dividing 2 on both sides, we get
\[ \Rightarrow x = \dfrac{{{y^2} + 2y + 5}}{2}\]
Rewriting the equation, we get
\[ \Rightarrow x = \dfrac{{{y^2} + 2y + 1}}{2} + \dfrac{4}{2}\]
Hence, using the identity \[{a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}\] in the numerator of first fraction in the RHS, we get,
\[ \Rightarrow x = \dfrac{{{{\left( {y + 1} \right)}^2}}}{2} + 2\]
Or \[x = \dfrac{1}{2}{\left( {y - \left( { - 1} \right)} \right)^2} + 2\]
Comparing this with \[x = m{\left( {y - b} \right)^2} + a\], where \[\left( {x,y} \right) = \left( {a,b} \right)\] is the vertex.
Thus, comparing the given equation with the general equation, we get,
Vertex \[\left( {x,y} \right) = \left( {2, - 1} \right)\]
Therefore, the vertex of the given parabola \[{y^2} + 2y - 2x + 5 = 0\] is \[\left( {2, - 1} \right)\]
Hence, this is the required answer.
Note:
A parabola is a curve having a focus and a directrix, such that each point on parabola is at equal distance from them. The axis of symmetry of a parabola is a line about which the parabola is symmetrical. When the parabola is vertical, the line of symmetry is vertical. When a quadratic function is graphed in the coordinate plane, the resulting parabola and corresponding axis of symmetry are vertical. Also, in this question, the parabola opens "sideways" and the axis of symmetry of the parabola is horizontal. The standard form of equation of a horizontal parabola is \[x = a{y^2} + by + c\] where \[a,b\] and \[c\] are all real numbers \[a \ne 0\].
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