
The vertex of the parabola \[{x^2} = 8y - 1\] is
A. \[\left( { - \dfrac{1}{8},0} \right)\]
B. \[\left( {\dfrac{1}{8},0} \right)\]
C. \[\left( {0,\dfrac{1}{8}} \right)\]
D. \[\left( {0, - \dfrac{1}{8}} \right)\]
Answer
233.1k+ views
Hint: First, we will use the standard equation of the parabola is \[{X^2} = 4aY\], where \[a\] is any real number and then the fact that the vertex \[\left( {X,Y} \right)\] of the standard equation of the parabola \[{X^2} = 4aY\] is \[\left( {0,0} \right)\], that is, we have \[\left( {X,Y} \right) = \left( {0,0} \right)\].
Apply these, and then use the given conditions to find the required value.
Complete step-by-step solution
We are given that the equation of the parabola is \[{x^2} = 8y - 1\].
Rewriting the above equation of the parabola by taking 8 common from the right hand side, we get
\[ \Rightarrow {x^2} = 8\left( {y - \dfrac{1}{8}} \right){\text{ .......eq.(1)}}\]
We know that the standard equation of the parabola is \[{X^2} = 4aY\], where \[a\] is any real number.
We will now compare the standard equation of the parabola with the given equation\[(1)\], we get
\[X = x\]
\[Y = y - \dfrac{1}{8}\]
\[a = 2\]
We know that the vertex \[\left( {X,Y} \right)\] of the standard equation of the parabola \[{X^2} = 4aY\] is \[\left( {0,0} \right)\], so we have \[\left( {X,Y} \right) = \left( {0,0} \right)\].
Substituting the values of \[X\] and \[Y\] in the above equation for vertex, we get
\[ \Rightarrow \left( {x,y - \dfrac{1}{8}} \right) = \left( {0,0} \right)\]
Adding the \[y\] coordinates of the above equation with \[\dfrac{1}{8}\] on each of the sides, we get
\[
\Rightarrow \left( {x,y} \right) = \left( {0,0 + \dfrac{1}{8}} \right) \\
\Rightarrow \left( {x,y} \right) = \left( {0,\dfrac{1}{8}} \right) \\
\]
Thus, the vertex of the given equation of the parabola is \[\left( {0,\dfrac{1}{8}} \right)\].
Hence, the option C is correct.
Note: In solving these types of questions, you should be familiar with the concept of the standard equation of the parabola and its vertex. We can also solve this question by taking the standard equation of the parabola is \[{\left( {X - h} \right)^2} = 4a\left( {Y - k} \right)\], where \[a\] is any real number and \[\left( {h,k} \right)\] is the coordinate of vertex. So we will have \[\left( {h,k} \right) = \left( {0,\dfrac{1}{8}} \right)\] in the equation \[(1)\], so we can say that the vertex is \[\left( {0,\dfrac{1}{8}} \right)\]. But this method has really few steps, which is helpful for competitive exams.
Apply these, and then use the given conditions to find the required value.
Complete step-by-step solution
We are given that the equation of the parabola is \[{x^2} = 8y - 1\].
Rewriting the above equation of the parabola by taking 8 common from the right hand side, we get
\[ \Rightarrow {x^2} = 8\left( {y - \dfrac{1}{8}} \right){\text{ .......eq.(1)}}\]
We know that the standard equation of the parabola is \[{X^2} = 4aY\], where \[a\] is any real number.
We will now compare the standard equation of the parabola with the given equation\[(1)\], we get
\[X = x\]
\[Y = y - \dfrac{1}{8}\]
\[a = 2\]
We know that the vertex \[\left( {X,Y} \right)\] of the standard equation of the parabola \[{X^2} = 4aY\] is \[\left( {0,0} \right)\], so we have \[\left( {X,Y} \right) = \left( {0,0} \right)\].
Substituting the values of \[X\] and \[Y\] in the above equation for vertex, we get
\[ \Rightarrow \left( {x,y - \dfrac{1}{8}} \right) = \left( {0,0} \right)\]
Adding the \[y\] coordinates of the above equation with \[\dfrac{1}{8}\] on each of the sides, we get
\[
\Rightarrow \left( {x,y} \right) = \left( {0,0 + \dfrac{1}{8}} \right) \\
\Rightarrow \left( {x,y} \right) = \left( {0,\dfrac{1}{8}} \right) \\
\]
Thus, the vertex of the given equation of the parabola is \[\left( {0,\dfrac{1}{8}} \right)\].
Hence, the option C is correct.
Note: In solving these types of questions, you should be familiar with the concept of the standard equation of the parabola and its vertex. We can also solve this question by taking the standard equation of the parabola is \[{\left( {X - h} \right)^2} = 4a\left( {Y - k} \right)\], where \[a\] is any real number and \[\left( {h,k} \right)\] is the coordinate of vertex. So we will have \[\left( {h,k} \right) = \left( {0,\dfrac{1}{8}} \right)\] in the equation \[(1)\], so we can say that the vertex is \[\left( {0,\dfrac{1}{8}} \right)\]. But this method has really few steps, which is helpful for competitive exams.
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