Question

How do you find parabola with equation $ y=a{{x}^{2}}+bx $ whose tangent line at (1,1) has equation $ y=3x-2 $ ?

Answer 1

Hint: This type of question is based on the concept of tangent lines and derivatives. As we know that the slope of the tangent line at a point on the function is equal to the derivative of the function at the same point. Therefore we will use this concept to find the values of a and b by forming two equations and solving them simultaneously. After finding the values of a and b we will substitute these values in the equation $ y=a{{x}^{2}}+bx $ to find the equation of the parabola.

Complete step by step answer:
It is given in the question that the parabola $ y=a{{x}^{2}}+bx $ and the tangent line whose equation is $ y=3x-2 $ meet at a common point (1,1).
Therefore,
The slope of the tangent at (1,1) is the derivative of $ y=a{{x}^{2}}+bx $ at x = 1.
Now, we will differentiate the equation $ y=a{{x}^{2}}+bx $ with respect to x, we get
 $ \dfrac{dy}{dx}=2ax+b $
Therefore the slope of the tangent at (1,1) is $ \dfrac{dy}{dx}=2a+b $ at x = 1
It is given that the equation of the tangent at (1,1) is $ y=3x-2 $
Now to find the slope of $ y=3x-2 $ , we use slope-intercept form $ y=mx+c $
Now comparing the tangent equation $ y=3x-2 $ with $ y=mx+c $ , we get value of m, which is the slope of the tangent.
After comparing we get
m =3
Therefore, the slope of $ y=3x-2 $ is 3
Therefore, we can write $ 2a+b=3 $ ………(1)
And we have been given that the point (1,1) lies on the parabola $ y=a{{x}^{2}}+bx $
Therefore, we get $ 1=a{{(1)}^{2}}+b(1)\Rightarrow a+b=1 $ …….(2)
Now from equation (2), we get
 $ a=1-b $
Now substituting the value of a in equation (1),we get
 $ \begin{align}
  & 2(1-b)+b=3 \\
 & \\
\end{align} $
On further simplifying, we get
 $ \begin{align}
  & \Rightarrow 2-2b+b=3 \\
 & \Rightarrow -b=3-2 \\
 & \Rightarrow b=-1 \\
 & \\
\end{align} $ $ $
Now substituting value of b in the equation $ a=1-b $ , we get
 $ \begin{align}
  & a=1-b \\
 & \Rightarrow a=1-(-1) \\
 & \Rightarrow a=2 \\
\end{align} $
Therefore, the equation of parabola is $ y=2{{x}^{2}}-x $ .

Note:
 These type of questions are usually based on concepts and calculations both. If the concepts are not clear then one can not even start solving them. In this question, the main point to be noted is that the slope of the tangent line at a point on the function is equal to the derivative of the function at the same point