
The axis of the parabola ${{x}^{2}}-4x-3y+10=0$ is
A.$y+2=0$
B.$x+2=0$
C.$y-2=0$
D.$x-2=0$
Answer
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Hint: Here, by adding and subtracting 4 to ${{x}^{2}}-4x-3y+10=0$, we can rewrite the equation as ${{(x-2)}^{2}}-3y+6=0$ Now, we have to convert the equation ${{(x-2)}^{2}}-3y+6=0$ into the vertex form $y={{\left( x-h \right)}^{2}}+k$, then the axis of symmetry will be $x-h=0$.
Complete step by step answer:
Here, we have to find the axis of the parabola ${{x}^{2}}-4x-3y+10=0$.
We know that every parabola has an axis of symmetry, which is the line that divides the parabola into two equal halves.
If our equation is in the standard form, $y=a{{x}^{2}}+bx+c$ then the formula for the axis of symmetry is:
$x=\dfrac{-b}{2a}$
If our equation is in the vertex form $y={{\left( x-h \right)}^{2}}+k$, then the formula for the axis of symmetry is:
$x=h$ or $x-h=0$.
Now, consider the equation
${{x}^{2}}-4x-3y+10=0$
We can add and subtract 4 to the above equation to get a perfect square. Hence, we will get:
${{x}^{2}}-4x+4-4-3y+10=0$
From the above equation we can say that ${{x}^{2}}-4x+4$ is the expansion of ${{\left( x-2 \right)}^{2}}$. Therefore, we can write:
${{(x-2)}^{2}}-4-3y+10=0$
Next, by addition we obtain:
${{(x-2)}^{2}}-3y+6=0$
In the next step, take $-3y$ to the right side, it becomes $3y$. Hence, we will obtain:
${{(x-2)}^{2}}-6=3y$.
The above equation is in the vertex form $y={{\left( x-h \right)}^{2}}+k$ then the formula then the formula for the axis of symmetry is:
$x-h=0$
Here, $h=2$, therefore, we can write:
$x-2=0$
Hence, we can say that the axis of the parabola ${{x}^{2}}-4x-3y+10=0$ is $x-2=0$.
Therefore, the correct answer for this question is option (d).
Note: Here, we can also find the axis of symmetry by converting the equation into the standard form or quadratic form $y=a{{x}^{2}}+bx+c$, then the axis of symmetry will be $x=\dfrac{-b}{2a}$. For that first, take all the terms except $3y$ to the right side then you will get the equation $3y={{x}^{2}}-4x+10$. Now dividing the equation by 3 you will get a quadratic equation and from the equation you can find an axis of symmetry.
Complete step by step answer:
Here, we have to find the axis of the parabola ${{x}^{2}}-4x-3y+10=0$.
We know that every parabola has an axis of symmetry, which is the line that divides the parabola into two equal halves.
If our equation is in the standard form, $y=a{{x}^{2}}+bx+c$ then the formula for the axis of symmetry is:
$x=\dfrac{-b}{2a}$
If our equation is in the vertex form $y={{\left( x-h \right)}^{2}}+k$, then the formula for the axis of symmetry is:
$x=h$ or $x-h=0$.
Now, consider the equation
${{x}^{2}}-4x-3y+10=0$
We can add and subtract 4 to the above equation to get a perfect square. Hence, we will get:
${{x}^{2}}-4x+4-4-3y+10=0$
From the above equation we can say that ${{x}^{2}}-4x+4$ is the expansion of ${{\left( x-2 \right)}^{2}}$. Therefore, we can write:
${{(x-2)}^{2}}-4-3y+10=0$
Next, by addition we obtain:
${{(x-2)}^{2}}-3y+6=0$
In the next step, take $-3y$ to the right side, it becomes $3y$. Hence, we will obtain:
${{(x-2)}^{2}}-6=3y$.
The above equation is in the vertex form $y={{\left( x-h \right)}^{2}}+k$ then the formula then the formula for the axis of symmetry is:
$x-h=0$
Here, $h=2$, therefore, we can write:
$x-2=0$
Hence, we can say that the axis of the parabola ${{x}^{2}}-4x-3y+10=0$ is $x-2=0$.
Therefore, the correct answer for this question is option (d).
Note: Here, we can also find the axis of symmetry by converting the equation into the standard form or quadratic form $y=a{{x}^{2}}+bx+c$, then the axis of symmetry will be $x=\dfrac{-b}{2a}$. For that first, take all the terms except $3y$ to the right side then you will get the equation $3y={{x}^{2}}-4x+10$. Now dividing the equation by 3 you will get a quadratic equation and from the equation you can find an axis of symmetry.
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