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.$\left( { - \dfrac{1}{2}, - \dfrac{1}{2}} \right)$
B.$\left( {\dfrac{1}{2},\dfrac{1}{2}} \right)$ $\left( {\dfrac{1}{2},\dfrac{1}{2}} \right)$
C.$\left( {\dfrac{1}{2}, - \dfrac{1}{3}} \right)$
D.$\left( {\dfrac{1}{2},\dfrac{1}{3}} \right)$

Answer 1

Hint : We simply need to find the equation of the normal which passes through the intersection point of the curve $y\left( {x - 2} \right)\left( {x - 3} \right) = x + 6$ and y-axis. We can then check which point as mentioned in the options satisfies the equation.

Complete step-by-step answer :
The equation of the curve is given by $y\left( {x - 2} \right)\left( {x - 3} \right) = x + 6$ and it intersects the y-axis.
Let us find the point of intersection of the given curve and the y-axis. Let this point of intersection is $\left( {0,\alpha } \right)$. Since $\left( {0,\alpha } \right)$ lies on the given curve, hence it will satisfy the equation of the curve.
Therefore,

$$\alpha \left( {0 - 2} \right)\left( {0 - 3} \right) = 0 + 6 \\\ \Rightarrow \alpha \left( { - 2} \right)\left( { - 3} \right) = 6 \\\ \Rightarrow \alpha = 1 \\\$$

Hence, the curve intersects the y-axis at point$\left( {0,1} \right)$.
Now, we need to find the equation of normal to curve at point. We first find the slope of tangent at point$\left( {0,1} \right)$.

$$\because y\left( {x - 2} \right)\left( {x - 3} \right) = x + 6 \\\ \Rightarrow y\left( {{x^2} - 5x + 6} \right) = x + 6 \\\$$

Differentiating both sides with respect to x, we get $\left( {0,1} \right)$ is $1$. We can now use slope point form to find the equation of the normal.
Putting $y = 1$ and $x = 0$ in above equation, we have

$$1\left( {0 - 5} \right) + \left( {0 - 0 + 6} \right){\left( {\dfrac{{dy}}{{dx}}} \right)_{0,1}} = 1 \\\ \Rightarrow - 5 + 6{\left( {\dfrac{{dy}}{{dx}}} \right)_{0,1}} = 1 \\\ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{0,1}} = \dfrac{6}{6} \\\ \therefore {\left( {\dfrac{{dy}}{{dx}}} \right)_{0,1}} = 1 \\\$$

Slope of tangent X slope of normal=$- 1$
Slope of normal$= - \dfrac{1}{{{\text{slope of tangent}}}}$

$$= \dfrac{{ - 1}}{{ - 1}} \\\ = 1 \\\$$

Thus the slope of normal which passes through $\left( {0,1} \right)$ is$1$. We can now use slope- point form to find the equation of the normal.

$$\dfrac{{y - 1}}{{x - 0}} = 1 \\\ \Rightarrow y + x = 1 \\\$$

Clearly, only option (c) satisfies the above equation of normal
So, the correct answer is “Option C”.

Note : The student must keep in mind that the x-coordinate of any point on y-axis is always 0. It is also very important to remember the relation between the slopes of tangent and normal at a point. Any point which lies on a curve must satisfy the equation of the curve.