Question

Given, $y=3$, $y=a{x}^2+b$
In the system of equations above, $a$ and $b$ are constants. For which of the following values of $a$ and $b$ does the system of equations have exactly two real solutions?

• A. $a=2,b=2$
• B. $a=2,b=4$
• C. $a=2,b=3$
• D. $a=4,b=3$

Answer 1

The correct option is A $a=2,b=2$

On substituting value of $y=3$ in second equation, we get
$a{x}^2+b=3$
$\Rightarrow a{x}^2+b-3=0$
$\Rightarrow a{x}^2=3-b$
$\Rightarrow x^2=\frac{3-b}{a}$
Since $x^2$ is positive quantity, therefore just $a=2$ and $b=2$ satisfies this.
Hence, option A is correct.