To find the value of k, you need to first find the gradient of the tangent of the curve. Then you will differentiate the equation of the curve and equate it to the gradient of the tangent to find the x-coordinate. Next you will substitute the x coordinate into the equation of the curve to find the y coordinate. Lastly you will substitute both the x and y coordinates into the linear equation (straight line equation) and find k.

Step 1: Find the gradient of the tangent from the equation of normal.

\displaystyle 3y+x=k

\displaystyle 3y=-x+k

\displaystyle y=-\frac{1}{3}x+\frac{k}{3}

\displaystyle {{m}_{2}}=-\frac{1}{3}

\displaystyle \text{Since}\ {{\text{m}}_{1}}\times {{m}_{2}}=-1

\displaystyle {{\text{m}}_{1}}=3

Step 2: Differentiate the equation of the curve, equate it to the gradient of the tangent to find x.

\displaystyle y={{x}^{2}}-5x+11

\displaystyle \frac{{dy}}{{dx}}=2x-5

\displaystyle 2x-5=3

\displaystyle 2x=8

\displaystyle x=4

Step 3: Substitute the x value into the equation of the curve to find the y coordinate. 

\displaystyle y={{4}^{2}}-5(4)+11

\displaystyle y=7

Step 4: Substitute both the x and y coordinate into the linear equation (straight line equation) to find k. 

\displaystyle 3y+x=k

\displaystyle 3(7)+4=k

\displaystyle k=25

 

At first sight, the question look very simple. But if you look a little deeper, you realize that it is not that easy to find the  x and y coordinates in order to find k. This is a typical exam of a O-Level exam question where you need to use formulas and methods from more then one topic to solve a problem. Some question will require knowledge from up to three different topics. So please don’t have a ‘tunnel vision’ and be flexible in using different methods and formulas.