Question

If the line $y=x$ is a tangent to the parabola $y=a{x}^2+bx+c$ at the point $(1,1)$ and the curve passes through $(-1,0)$, then

• A. $a=b=-1,c=3$
• B. $a=b=\frac{1}{2},c=0$
• C. $a=c=\frac{1}{4},b=\frac{1}{2}$
• D. $a=0,b=c=\frac{1}{2}$

Answer 1

The correct option is C $a=c=\frac{1}{4},b=\frac{1}{2}$

$y=x$ is a tangent
$\therefore$slopes are equal.
$\frac{dy}{dx}=2ax+b$
$\Rightarrow 1=2a+b$ at $(1,1)\cdots \left(1\right)$
Also, the parabola passes through $(1,1)$
$\Rightarrow a+b+c=1\cdots \left(2\right)$
The parabola passes through $(-1,0)$
$\Rightarrow 0=a-b+c\cdots \left(3\right)$
Solving $\left(1\right),\left(2\right),\left(3\right)$, we get -
$\therefore a=c=\frac{1}{4}$
and $b=\frac{1}{2}$