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Question: 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.\...

The axis of the parabola x24x3y+10=0{{x}^{2}}-4x-3y+10=0 is
A.y+2=0y+2=0
B.x+2=0x+2=0
C.y2=0y-2=0
D.x2=0x-2=0

Explanation

Solution

Hint: Here, by adding and subtracting 4 to x24x3y+10=0{{x}^{2}}-4x-3y+10=0, we can rewrite the equation as (x2)23y+6=0{{(x-2)}^{2}}-3y+6=0 Now, we have to convert the equation (x2)23y+6=0{{(x-2)}^{2}}-3y+6=0 into the vertex form y=(xh)2+ky={{\left( x-h \right)}^{2}}+k, then the axis of symmetry will be xh=0x-h=0.
Complete step by step answer:
Here, we have to find the axis of the parabola x24x3y+10=0{{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=ax2+bx+cy=a{{x}^{2}}+bx+c then the formula for the axis of symmetry is:
x=b2ax=\dfrac{-b}{2a}
If our equation is in the vertex form y=(xh)2+ky={{\left( x-h \right)}^{2}}+k, then the formula for the axis of symmetry is:
x=hx=h or xh=0x-h=0.
Now, consider the equation
x24x3y+10=0{{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:
x24x+443y+10=0{{x}^{2}}-4x+4-4-3y+10=0
From the above equation we can say that x24x+4{{x}^{2}}-4x+4 is the expansion of (x2)2{{\left( x-2 \right)}^{2}}. Therefore, we can write:
(x2)243y+10=0{{(x-2)}^{2}}-4-3y+10=0
Next, by addition we obtain:
(x2)23y+6=0{{(x-2)}^{2}}-3y+6=0
In the next step, take 3y-3y to the right side, it becomes 3y3y. Hence, we will obtain:
(x2)26=3y{{(x-2)}^{2}}-6=3y.
The above equation is in the vertex form y=(xh)2+ky={{\left( x-h \right)}^{2}}+k then the formula then the formula for the axis of symmetry is:
xh=0x-h=0
Here, h=2h=2, therefore, we can write:
x2=0x-2=0
Hence, we can say that the axis of the parabola x24x3y+10=0{{x}^{2}}-4x-3y+10=0 is x2=0x-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=ax2+bx+cy=a{{x}^{2}}+bx+c, then the axis of symmetry will be x=b2ax=\dfrac{-b}{2a}. For that first, take all the terms except 3y3y to the right side then you will get the equation 3y=x24x+103y={{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.