Question
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 x2−4x−3y+10=0 is
A.y+2=0
B.x+2=0
C.y−2=0
D.x−2=0
Solution
Hint: Here, by adding and subtracting 4 to x2−4x−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=(x−h)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 x2−4x−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+c then the formula for the axis of symmetry is:
x=2a−b
If our equation is in the vertex form y=(x−h)2+k, then the formula for the axis of symmetry is:
x=h or x−h=0.
Now, consider the equation
x2−4x−3y+10=0
We can add and subtract 4 to the above equation to get a perfect square. Hence, we will get:
x2−4x+4−4−3y+10=0
From the above equation we can say that x2−4x+4 is the expansion of (x−2)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=(x−h)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 x2−4x−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=ax2+bx+c, then the axis of symmetry will be x=2a−b. For that first, take all the terms except 3y to the right side then you will get the equation 3y=x2−4x+10. Now dividing the equation by 3 you will get a quadratic equation and from the equation you can find an axis of symmetry.