Question

A curve has equation $y={{x}^{2}}-4x+4$ and a line has equation$y=mx$, where $m$ is a constant.
1. For the case where $m=1$, the curve and the line intersect at the points A and B. Find the coordinates of the midpoint of AB.
2. Find the non-zero value of $m$ for which the line is a tangent to the curve, and find the coordinates of the point where the tangent touches the curve.

Answer 1

In this type of question we have to use the concept of the midpoint formula as well as the equation of a tangent to the curve.
In the first case we have to put $m=1$ in the equation of line $y=mx$. Then we compare the equations of line and curve to obtain the points of intersection A and B, after that by using the midpoint formula we can find the coordinates of the midpoint of AB. We know that the coordinates of the midpoint $\left( {{x}_{m}},{{y}_{m}} \right)$ of the line joining the points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$are given by, $\left( {{x}_{m}},{{y}_{m}} \right)=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$.
In the second case, first we find the slope of the line and the curve which we can obtain by finding the derivative $\dfrac{dy}{dx}$ and by comparing both slopes we will get the required result.

Complete step by step answer:
1. We have to find the coordinates of the midpoint of AB. Where A and B are the points of intersection of the curve having equation $y={{x}^{2}}-4x+4$ and a line whose equation is $y=mx$, where $m$ is a constant and $m=1$.
As we have given that $m=1$ so that the equation of line becomes, $y=x$.
Now, as A and B are the points of intersection of the line $y=x$ and the curve $y={{x}^{2}}-4x+4$

& \Rightarrow {{x}^{2}}-4x+4=x \\\ & \Rightarrow {{x}^{2}}-4x+4-x=0 \\\ & \Rightarrow {{x}^{2}}-x-4x+4=0 \\\ & \Rightarrow x\left( x-1 \right)-4\left( x-1 \right)=0 \\\ & \Rightarrow \left( x-1 \right)\left( x-4 \right)=0 \\\ & \Rightarrow x=1,4 \\\ \end{aligned}$$ Now as we have $$y=x$$ hence we will get, $$\Rightarrow \text{When }x=1,y=1\text{ and when }x=4,y=4$$ Thus the points of intersection of the line $$y=x$$ and the curve $$y={{x}^{2}}-4x+4$$ are $$A\left( 1,1 \right)\And B\left( 4,4 \right)$$. Now, to find the coordinates of the midpoint of AB we have to use midpoint formula. We know that the coordinates of the midpoint $$\left( {{x}_{m}},{{y}_{m}} \right)$$ of the line joining the points $$\left( {{x}_{1}},{{y}_{1}} \right)$$ and $$\left( {{x}_{2}},{{y}_{2}} \right)$$are given by, $$\left( {{x}_{m}},{{y}_{m}} \right)=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$$. Thus the midpoint of AB where $$A\equiv \left( 1,1 \right)\And B\equiv \left( 4,4 \right)$$ $$\begin{aligned} & \Rightarrow \left( {{x}_{m}},{{y}_{m}} \right)=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right) \\\ & \Rightarrow \left( {{x}_{m}},{{y}_{m}} \right)=\left( \dfrac{1+4}{2},\dfrac{1+4}{2} \right) \\\ & \Rightarrow \left( {{x}_{m}},{{y}_{m}} \right)=\left( \dfrac{5}{2},\dfrac{5}{2} \right) \\\ \end{aligned}$$  2\. Now we have to find the non-zero value of $$m$$ for which the line $$y=mx$$ is a tangent to the curve $$y={{x}^{2}}-4x+4$$, and also we have to find the coordinates of the point where the tangent touches the curve. Let us first find the tangent to the curve $$y={{x}^{2}}-4x+4$$ and to the line $$y=mx$$. As we know that the tangent can be obtained by taking the derivative with respect to x, hence $$\begin{aligned} & \Rightarrow \text{Tangent to the curve }\left( y={{x}^{2}}-4x+4 \right)=\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{2}}-4x+4 \right) \\\ & \Rightarrow \text{Tangent to the curve }\left( y={{x}^{2}}-4x+4 \right)=\dfrac{dy}{dx}=2x-4\cdots \cdots \cdots \left( i \right) \\\ & \Rightarrow \text{Tangent to the line }\left( y=mx \right)=\dfrac{dy}{dx}=\dfrac{d}{dx}\left( mx \right) \\\ & \Rightarrow \text{Tangent to the line }\left( y=mx \right)=\dfrac{dy}{dx}=m\cdots \cdots \cdots \left( ii \right) \\\ \end{aligned}$$ On comparing the equation $$\left( i \right)$$ and $$\left( ii \right)$$ we get, $$\Rightarrow m=\left( 2x-4 \right)\cdots \cdots \cdots \left( iii \right)$$ Now by substituting this value of $$m$$ in equation of line, we can write, $$\begin{aligned} & \Rightarrow y=\left( 2x-4 \right)x \\\ & \Rightarrow y=2{{x}^{2}}-4x\cdots \cdots \cdots \left( iv \right) \\\ \end{aligned}$$ But from the equation of curve we have $$y={{x}^{2}}-4x+4$$, $$\begin{aligned} & \Rightarrow {{x}^{2}}-4x+4=2{{x}^{2}}-4x \\\ & \Rightarrow {{x}^{2}}+4=2{{x}^{2}} \\\ & \Rightarrow 4={{x}^{2}} \\\ & \Rightarrow \pm 2=x \\\ & \Rightarrow x=2,-2 \\\ \end{aligned}$$ Hence, from equation $$\left( iv \right)$$ we can write, $$\Rightarrow \text{When }x=2,y=0\text{ and when }x=-2,y=16$$ Also from equation $$\left( iii \right)$$ we can obtain values for $$m$$ $$\Rightarrow \text{When }x=2,m=0\text{ and when }x=-2,m=-8$$ Therefore, the non-zero value of $$m$$ is $$-8$$ for which the line $$y=mx$$ is a tangent to the curve $$y={{x}^{2}}-4x+4$$, and also the coordinates of the point where the tangent touches the curve are $$\left( -2,16 \right)$$.  **Note:** In the first part of the question students have to remember that to find the points of intersection of the line $$y=x$$ and the curve $$y={{x}^{2}}-4x+4$$ by equating the two equations. Also students have to note that they have to use the midpoint formula to obtain the midpoint of AB. In the second part of the question students have to remember that the tangent of the curve can be obtained by taking the derivative with respect to x. Also students have to note that the required value of m must be non-negative so from the obtained values of m students have to accept the value -8.