Question

If tangent to the curve $ y = 6x - {x^2} $ is parallel to the line $ 4x - 2y - 1 = 0 $ , then the point of tangency on the curve is
A. $ \left( {2,8} \right) $
B. $ \left( {8,2} \right) $
C. $ \left( {6,1} \right) $
D. $ \left( {4,2} \right) $

Answer 1

Hint: First, we shall analyze the given information so that we are able to solve the given problem. Here, it is given that the tangent to the curve is parallel to the line. We need to find the tangent to the curve $ y = 6x - {x^2} $ and compare it with the slope of the line $ 4x - 2y - 1 = 0 $
Then we will obtain the desired solution.
Formula to be used:
a) The formula to obtain the power rule of the derivative is as follows.
 $ \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}} $
b) The formula to obtain the slope of a line \[ax + by + c = 0\] is as follows.
The slope of a line\[ax + by + c = 0\], $ m = \dfrac{{ - a}}{b} $ where $ a $ is the x-coordinate and $ b $ is the y-coordinate.

Complete step by step answer:
It is given that $ y = 6x - {x^2} $
The tangent to curve to the curve $ y = 6x - {x^2} $ is obtained by calculating the derivative of $ y $ .
Hence, $ \dfrac{{dy}}{{dx}} = 6{x^{1 - 1}} - 2{x^{2 - 1}} $ (Here we applied the power rule of derivative $ \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}} $ )
That is $ \dfrac{{dy}}{{dx}} = 6 - 2x $ is the required tangent.
And, it is given that the above tangent is parallel to the line $ 4x - 2y - 1 = 0 $
Now, we shall find the slope of the line $ 4x - 2y - 1 = 0 $
The formula to obtain the slope of a line \[ax + by + c = 0\] is as follows.
The slope of a line\[ax + by + c = 0\], $ m = \dfrac{{ - a}}{b} $ where $ a $ is the x-coordinate and $ b $ is the y-coordinate.
Hence, the slope of a line $ 4x - 2y - 1 = 0 $ , $ m = \dfrac{{ - 4}}{{ - 2}} $
 $ \Rightarrow m = 2 $
Hence $ \dfrac{{dy}}{{dx}} = 2 $
Since it is given that the tangent is parallel to the line $ 4x - 2y - 1 = 0 $ , we need to compare both $ \dfrac{{dy}}{{dx}} $
That is, we need to compare $ \dfrac{{dy}}{{dx}} = 6 - 2x $ and $ \dfrac{{dy}}{{dx}} = 2 $
 $ 6 - 2x = 2 $
 $ 6 - 2 = 2x $
 $ \Rightarrow 2x = 4 $
 $ \Rightarrow x = 2 $
Now, we shall substitute $ x = 2 $ in $ y = 6x - {x^2} $
 $ y = 6 \times 2 - {\left( 2 \right)^2} $
 $ y = 12 - 4 $
 $ y = 8 $
Hence, we get $ x = 2 $ and $ y = 8 $ .
Therefore, the required point of tangency will be $ \left( {2,8} \right) $

So, the correct answer is “Option A”.

Note: If we are given that the tangent is parallel to the curve, then we need to find the derivative of the given curve. To find the derivative of the given curve, we have applied the power rule of the derivative here. Then, we have compared the resultant derivative with the slope of the line $ 4x - 2y - 1 = 0 $ Therefore, the required point of tangency will be $ \left( {2,8} \right) $ .