Question

Equation of tangent to the curve \[y\left( x-2 \right)\left( x-3 \right)-x+7=0\] at a point where the curve cuts the X – axis.
(a)\[x-20y=7\]
(b) \[x-20y+7=0\]
(c) \[x+20y-7=0\]
(d) \[x+20y+7=0\]

Answer 1

Hint: We solve this problem by finding the coordinates of point where the given curve touches the X-axis.
We use the condition that when any curve touches the X – axis then \[y=0\]
We substitute \[y=0\] in given curve equation to find the co – ordinates of point.
Then we find the slope at that point by using the formula that slope of any curve is given as
\[m=\left( \dfrac{dy}{dx} \right)\]
We use the standard formula of differentiation that is
\[\dfrac{d}{dx}\left( f\left( x \right)\times g\left( x \right)\times h\left( x \right) \right)={f}'\left( x \right)\times g\left( x \right)\times h\left( x \right)+f\left( x \right)\times {g}'\left( x \right)\times h\left( x \right)+f\left( x \right)\times g\left( x \right)\times {h}'\left( x \right)\]
Then we substitute the point in the slope equation to get the slope.
We have a standard result of finding the line equation that is the equation of line passing through point \[\left( a.b \right)\] and have a slope \[m\] is given as
\[\left( y-b \right)=m\left( x-a \right)\]

Complete step by step answer:
We are given that the equation of curve as
\[\Rightarrow y\left( x-2 \right)\left( x-3 \right)-x+7=0\]
Let us assume a rough figure of the given curve hat touches X – axis as

We are asked to find the tangent where the curve touches the X – axis.
We know that the condition that when any curve touches the X – axis then \[y=0\]
Now, by substituting the value \[y=0\] in the given equation we get
\[\begin{align}
  & \Rightarrow 0\left( x-2 \right)\left( x-3 \right)-x+7=0 \\
 & \Rightarrow -x+7=0 \\
 & \Rightarrow x=7 \\
\end{align}\]
So, we can say that co – ordinates of point where the given curve touches the X – axis is \[\left( 7,0 \right)\]
Now, let us find the slope of the curve at \[\left( 7,0 \right)\]
Let us take the given curve and differentiate the equation with respect to \[x\] we get
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( y\left( x-2 \right)\left( x-3 \right)-x+7 \right)=\dfrac{d}{dx}\left( 0 \right) \\
 & \Rightarrow \dfrac{d}{dx}\left( y\left( x-2 \right)\left( x-3 \right) \right)-\dfrac{dx}{dx}+\dfrac{d}{dx}\left( 7 \right)=0 \\
\end{align}\]
We know that the standard formula of differentiation that is
\[\dfrac{d}{dx}\left( f\left( x \right)\times g\left( x \right)\times h\left( x \right) \right)={f}'\left( x \right)\times g\left( x \right)\times h\left( x \right)+f\left( x \right)\times {g}'\left( x \right)\times h\left( x \right)+f\left( x \right)\times g\left( x \right)\times {h}'\left( x \right)\]

By using this formula to above equation we get
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}\left( x-2 \right)\left( x-3 \right)+y\left( 1 \right)\left( x-3 \right)+y\left( x-2 \right)\left( 1 \right)-1+0=0 \\
 & \Rightarrow \dfrac{dy}{dx}\left( x-2 \right)\left( x-3 \right)+y\left( x-3 \right)+y\left( x-2 \right)-1=0 \\
\end{align}\]
Now, by substituting the point \[\left( 7,0 \right)\] in above equation we get
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}\left( 7-2 \right)\left( 7-3 \right)+0\left( 7-3 \right)+0\left( 7-2 \right)-1=0 \\
 & \Rightarrow \dfrac{dy}{dx}\times 5\times 4=1 \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{20} \\
\end{align}\]
Let us assume that the slope of curve at \[\left( 7,0 \right)\] as \[m\]
We know that the slope of any curve is given as
\[m=\left( \dfrac{dy}{dx} \right)\]
By using this condition we get the slope of curve at \[\left( 7,0 \right)\] as
\[\Rightarrow m=\dfrac{1}{20}\]
We know that the equation of line passing through point \[\left( a.b \right)\] and have a slope \[m\] is given as
\[\left( y-b \right)=m\left( x-a \right)\]
By using the above formula we get the required equation of line as
\[\begin{align}
  & \Rightarrow y-0=\dfrac{1}{20}\left( x-7 \right) \\
 & \Rightarrow 20y=x-7 \\
 & \Rightarrow x-20y=7 \\
\end{align}\]
Therefore, we can conclude that the equation of tangent o given curve is \[x-20y=7\]
So, option (a) is correct answer.
Note:
Students may do mistake in considering the coordinate of the point when a curve touches the X-axis.
We have the condition that the y coordinate of a point when a curve touches X-axis as \[y=0\]
But some students may misunderstand that it touches X – axis so \[x=0\]
But this is not correct because touching the X-axis means it can touch anywhere on the X – axis. So, there will be some value for x coordinate.