Question

If the curve $y=a{x}^2+bx+c;x\in R$ passes through the point $(1,2)$ and the tangent line to this curve at origin is $y=x$, then the possible values of $a,b,c$ are:

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

Answer 1

The correct option is A

$a=1,b=1,c=0$

Explanation for the correct option.

Step 1: Find the tangent.

Given that, the curve $y=a{x}^2+bx+c;x\in R$ passes through the point $(1,2)$ and the tangent line to this curve at origin is $y=x$.

As the curve passes through $(1,2)$ so the curve satisfies the point.

$\Rightarrow 2=a+b+c...\left(1\right)$

Now tangent to the curve can be found by differentiating the curve.

$\Rightarrow \frac{dy}{dx}=2ax+b$

On comparing $y=x$ with $y=mx+c;m$ is slope we get: $m=1$

$$\begin{gathered} \Rightarrow \frac{dy}{dx}{|}_{\left(0,0\right)}=1 \\ \Rightarrow b=1....\left(2\right) \end{gathered}$$

Step 2: Find the value of constants

Substituting equation (2) in (1) we get: $a+c=1$

Since, the curve passes through the origin.

So, $c=0$

$\Rightarrow a=1$

Therefore, $a=1,b=1,c=0$.

Hence, option A is correct.