Question

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 1

Answer: (D)

$x-2=0$

Method I:-

Equation of parabola-

$x^2-4x-3y+10=0$

${(x-2)}^2-4-3y+10=0$

$\Rightarrow {(x-2)}^2-3y+6=0$

$\Rightarrow 3y-6={(x-2)}^2$

For axis of parabola-

$x-2=0$

• Method II:-
  

Equation of parabola-

$x^2-4x-3y+10=0$

$3y=x^2-4x+10$

$y=\frac{1}{3}{x}^2-\frac{4}{3}x+\frac{10}{3}$

For a quadratic function $y=a{x}^2+bx+c$, axis of symmetry is a vertical line $x=-\frac{b}{2a}$

Therefore, for the given equation of parabola $y=\frac{1}{3}{x}^2-\frac{4}{3}x+\frac{10}{3}$,

$a=\frac{1}{3}$

$b=-\frac{4}{3}$

$c=\frac{10}{3}$

$\therefore$ Axis of symmetry will be-

$x=-\frac{b}{2a}=-\frac{(-\frac{4}{3})}{2\left(\frac{1}{3}\right)}$

$\Rightarrow x=2$

$\Rightarrow x-2=0$

Hence axis of parabola $x^2-4x-3y+10=0$ will be $x-2=0$.