Question
The normal to the curve $$y\left( {x - 2} \right)\left( {x - 3} \right) = x + 6$$ at the point where the curve intersects the $$y$$ -axis passes through the point:
A.
$$\left( {\frac{1}{2},\frac{1}{3}} \right)$$
B.
$$\left( { - \frac{1}{2}, - \frac{1}{2}} \right)$$
C.
$$\left( {\frac{1}{2},\frac{1}{2}} \right)$$
D.
$$\left( {\frac{1}{2}, - \frac{1}{3}} \right)$$
Answer :
$$\left( {\frac{1}{2},\frac{1}{2}} \right)$$
Solution :
We have $$y = \frac{{x + 6}}{{\left( {x - 2} \right)\left( {x - 3} \right)}}$$
At $$y$$ -axis, $$x = 0 \Rightarrow y = 1$$
On differentiating, we get
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{\left( {{x^2} - 5x + 6} \right)\left( 1 \right) - \left( {x + 6} \right)\left( {2x - 5} \right)}}{{{{\left( {{x^2} - 5x + 6} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = 1\,{\text{at}}\,{\text{point}}\left( {0,1} \right) \cr
& \therefore \,{\text{Slope}}\,{\text{of}}\,{\text{normal}} = - 1 \cr
& {\text{Now}}\,{\text{equation}}\,{\text{of}}\,{\text{normal}}\,{\text{is}}\,y - 1 = - 1\left( {x - 0} \right) \cr
& \Rightarrow y - 1 = - x \cr
& x + y = 1 \cr
& \therefore \,\,\left( {\frac{1}{2},\frac{1}{2}} \right)\,{\text{satisfy}}\,{\text{it}}. \cr} $$