Question

How do you find an equation of line tangent to the curve at the given point
 ${x^3} + {y^3} = 36xy$ , given point $\left( {18,18} \right)$ ?
A. $x + y = 18$
B. $x + y = 36$
C. $x + y = - 18$
D. $x + 2y = 18$

Answer 1

Hint: A tangent is a line that touches a curve at a single point and does not cross through it. The equation of a line is given by $y - {y_0} = m\left( {x - {x_0}} \right)$where ${y_0}\,and\,{x_0}$ are the points where tangent touches the curve and $m$ is the slope of line.
The slope of tangent $m$ can be found by differentiating the equation of the curve and putting the values of the given point of intersection in it. That is
 $m = \dfrac{{dy}}{{dx}}$ at $\left( {18,18} \right)$

Complete step-by-step answer:
 step1 write the given equation of curve
 ${x^3} + {y^3} = 36xy$
Step2 now differentiate both side of the equation with respect to $x$
$\dfrac{d}{{dx}}\left( {{x^3} + {y^3}} \right) = \dfrac{d}{{dx}}\left( {36xy} \right)$
For differentiating the two variable terms we will use the chain rule of differentiation
 $3{x^2} + 3{y^2}\dfrac{{dy}}{{dx}} = 36\left( {y\dfrac{d}{{dx}}x + x\dfrac{{dy}}{{dx}}} \right)$
Now simplifying the terms
 $3{x^2} + 3{y^2}\dfrac{{dy}}{{dx}} = 36\left( {y + x\dfrac{{dy}}{{dx}}} \right)$
Now separating the derivative terms we get
 $3{y^2}\dfrac{{dy}}{{dx}} - 36x\dfrac{{dy}}{{dx}} = 36y - 3{x^2}$
Taking $\dfrac{{dy}}{{dx}}$ common and transferring other terms on right side we get
 $\dfrac{{dy}}{{dx}} = \dfrac{{36y - 3{x^2}}}{{3{y^2} - 36x}}$ ……………… $\left( 1 \right)$
Now putting the given point of intersection in equation $1$
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {36 \times 18} \right) - \left( {3 \times {{18}^2}} \right)}}{{\left( {3 \times {{18}^2}} \right) - \left( {36 \times 18} \right)}}$
On simplifying we get the value of slope
 $ \Rightarrow m = \dfrac{{dy}}{{dx}} = - 1$
Now to find the equation of tangent we will put the value of slope and given point in standard equation of line $y - {y_0} = m\left( {x - {x_0}} \right)$ we get,
 $ \Rightarrow y - 18 = - 1\left( {x - 18} \right)$ …………..$\left( 2 \right)$
Separating variables and constant terms we get
 $ \Rightarrow x + y = 36$
This is the required solution of tangent to the given curve.
So, the correct answer is “Option B”.

Note: To check that our answer is correct or not, put the value of point in the equation of the line. If the point satisfies the equation of line then our answer is correct. Let us check our answer.
Put the value of point in equation 2 we get,
 $ \Rightarrow 18 - 18 = - 1\left( {18 - 18} \right)$
 On further solving
 $ \Rightarrow 0 = 0$
Since LHS $ = $ RHS our answer is correct.