Question

Solve the following equation for tangent on the curve at $(2,2)$:${x^2} - 2xy + {y^2} + 2x + y - 6 = 0$
(a) $2x + y + 6 = 0$
(b) $2x - y - 6 = 0$
(c) $2x + y - 6 = 0$
(d) None of the above

Answer 1

We will use the most eccentric concept of derivations. Considering the ‘y’ variable as a derivative agent or term the solution is solved by using laws of derivation like addition, subtraction, multiplication, dividation, etc. and using the straight line equation particularly. As a result, analysing the obtained values with the options provided, the required answer can be reached.

Complete step-by-step solution:
The given tangent equation on the curve is,
${x^2} - 2xy + {y^2} + 2x + y - 6 = 0$
Therefore, derivating the above given equation with respect to the ‘x’ variable, can reach up to a desire output,
$\Rightarrow {x^2} - 2xy + {y^2} + 2x + y - 6 = 0$
Now, hence derive the equation in terms of ‘$y$’ with respect to ‘$x$’, we get
$\Rightarrow 2x - 2\left[ {x\left( {\dfrac{{dy}}{{dx}}} \right) + y(1)} \right] + 2y\left( {\dfrac{{dy}}{{dx}}} \right) + 2 + \dfrac{{dy}}{{dx}} - 0 = 0$
… ($\because$Using the laws of derivation such as for multiplication and derivation of any constant is always zero, here that constant is $\dfrac{d}{{dx}}(6) = 0$ respectively)
As a result, solve the equation predominantly, we get
$\Rightarrow 2x - 2x\left( {\dfrac{{dy}}{{dx}}} \right) - 2y + 2y\left( {\dfrac{{dy}}{{dx}}} \right) + 2 + \dfrac{{dy}}{{dx}} = 0$
Taking $\dfrac{{dy}}{{dx}}$ common in one bracket and remaining terms in another bracket, we get
$\Rightarrow ( - 2x + 2y + 1)\left( {\dfrac{{dy}}{{dx}}} \right) + (2x - 2y + 2) = 0$
Again, separating all the terms from $\dfrac{{dy}}{{dx}}$, shifting the terms on right – hand side, we get

$$\Rightarrow ( - 2x + 2y + 1)\left( {\dfrac{{dy}}{{dx}}} \right) = - (2x - 2y + 2) \\\ \Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = - \left[ {\dfrac{{2x - 2y + 2}}{{ - 2x + 2y + 1}}} \right] \\\$$

But, here, we have given that curve exists on the points $(2,2)$ respectively,
Therefore, the equation becomes
$\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{at(2,2)}} = - \left[ {\dfrac{{2x - 2y + 2}}{{ - 2x + 2y + 1}}} \right]$
Substituting the given points that is $x = 2$ and $y = 2$ respectively, we get
$\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{at(2,2)}} = - \left[ {\dfrac{{2(2) - 2(2) + 2}}{{ - 2(2) + 2(2) + 1}}} \right]$
Primarily solving the equation, we get

\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{at(2,2)}} = - \left[ {\dfrac{{4 - 4 + 2}}{{ - 4 + 4 + 1}}} \right] \\\ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{at(2,2)}} = - \left( {\dfrac{2}{1}} \right) = - 2 \\\ $$ … (i) Now, since we know that, the straight line equation for the line (or, tangent) is, $y = mx + c$ … (ii) Where, $m$is slope or gradient of the line we have find in the above solution [from (i)], $c$ is the y-intercept that is the point on the line respectively. Hence, substituting all the required values that is $x = 2$,$y = 2$ and $m = {\left( {\dfrac{{dy}}{{dx}}} \right)_{at(2,2)}} = - 2$ in the equation (ii), we get $$ \Rightarrow 2 = - 2(2) + c = c - 4$$ $ \Rightarrow c = 2 + 4 = 6$ … (iii) Now, since from the given number of options for the respective question, option (c) satisfies the value/s [equation (iii)] that has been solved, Since, substituting $x = 2,y = 2$ and $c = 6$, we get $ \Rightarrow 2x + y - c = 0 = 2x + y - 6$ $ \Rightarrow 2(2) + 2 - 6 = 0 = 4 + 2 - 6$ $ \Rightarrow 0 = 0$ i.e. $L.H.S. = R.H.S.$ **Therefore, option (c) is correct.** **Note:** One must remember the concept of derivation, how to differentiate the equation with respect to which variable, etc.? Also, laws of derivation such as $\dfrac{d}{{dx}}(xy) = x\dfrac{{dy}}{{dx}} + y(1)$ is the multiplication rule of derivation used here. Derivation of any constant number is always zero. Deriving the equation with the same term is always one $\dfrac{d}{{dx}}(x) = 1$. Algebraic identities play a significant role in solving this problem.