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Question: The vertex of the parabola \({y^2} - 10y + x + 22 = 0\) is A. \((3,4)\) B. \((3,5)\) C. \((5,...

The vertex of the parabola y2−10y+x+22=0{y^2} - 10y + x + 22 = 0 is
A. (3,4)(3,4)
B. (3,5)(3,5)
C. (5,3)(5,3)
D. None of these

Explanation

Solution

Here we have to find the vertex of the given parabola. We will first simplify the equation and then solve it to the vertex form, then by comparing it with the general equation of the parabola, we will be able to find the required vertex of the parabola. The equation for a parabola can also be written in vertex form as x=m(y−b)2+ax = m{(y - b)^2} + a, where (a,b)(a,b) is the vertex of the parabola.

Complete step by step answer:
As per the question we have y2−10y+x+22=0{y^2} - 10y + x + 22 = 0.
It can be written as x=−y2+10y−22x = - {y^2} + 10y - 22.
We know the algebraic expression that (a−b)2=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2ab. So we can write the above equation as (y−5)2=y2+25−10y{(y - 5)^2} = {y^2} + 25 - 10y. We can divide the constant term in two parts and then write it in the above form.
So we can write it as x=−(y−5)2+3x = - {(y - 5)^2} + 3. Now this is in the vertex form.
By comparing it with the vertex form we can write (a,b)=(3,5)(a,b) = (3,5).

Hence the correct option is B.

Note: We should know that a parabola is a curve having a focus and a directrix, such that each point on parabola is equal distance from them. The standard form of the equation of a horizontal parabola is x=ay2+by+cx = a{y^2} + by + c, where a,ba,b and cc are the real numbers and a≠0a \ne 0. We should note that the axis of symmetry of a parabola is a line about which the parabola is symmetrical.