Question

How do you find horizontal and vertical tangent lines after using implicit differentiation \[{{x}^{2}}+xy+{{y}^{2}}=27\]?

Answer 1

Hint: First differentiate the whole function with respect to \[x\] and then collect all the \[dy/dx\] on the same side and solve for \[dy/dx\]. On finding \[dy/dx\] that is the differentiation of \[y\] with respect to \[x\] is the slope of the tangent at any point \[(x,y)\] on the given function \[y\]. Now put this \[dy/dx\] is equal to zero and not defined separately as the slope of a horizontal line is zero and for vertical is not defined then on simplifying, we would get the coordinates of those required points.

Complete step by step solution:
As it is given a function \[{{x}^{2}}+xy+{{y}^{2}}=27\]
Since it is implicit as it is a mixed variable function and we will not get \[y\] directly on knowing \[x\]
Now differentiating both sides with respect to \[x\]
\[\dfrac{d}{dx}({{x}^{2}}+xy+{{y}^{2}})=\dfrac{d}{dx}(27)\]
Since the differentiation of a constant term is zero
\[\Rightarrow \dfrac{d}{dx}({{x}^{2}}+xy+{{y}^{2}})=0\]
And we know that \[\dfrac{d}{dx}(f+g)=\dfrac{df}{dx}+\dfrac{dg}{dx}\]
\[\Rightarrow \dfrac{d}{dx}({{x}^{2}})+\dfrac{d}{dx}(xy)+\dfrac{d}{dx}({{y}^{2}})=0\]
Since the differentiation of a term of form \[{{x}^{n}}{{y}^{m}}\] is first differentiate \[{{y}^{m}}\] consider as \[{{x}^{m}}\] then multiply it with differentiation of \[y\] with respect to \[x\] then add the term with keeping function of \[y\] as it is and differentiate \[x\]
i.e. \[\dfrac{d({{x}^{n}}{{y}^{m}})}{dx}={{x}^{n}}m{{y}^{m-1}}\dfrac{dy}{dx}+{{y}^{m}}n{{x}^{n-1}}\]
Now differentiating with respect to \[x\]
\[\Rightarrow 2x+\left( x\dfrac{dy}{dx}+y \right)+2y\dfrac{dy}{dx}=0\]
On simplifying
\[\dfrac{dy}{dx}=\dfrac{-(2x+y)}{x+2y}\]
We know that \[\dfrac{dy}{dx}\] is the slope of the tangent at any point \[(x,y)\] on given function.
Ans it is given that the tangents are horizontal and vertical
As for horizontal tangent, \[\dfrac{dy}{dx}=0\]
\[\Rightarrow \dfrac{-(2x+y)}{x+2y}=0\]
For this the numerator must be zero
\[\Rightarrow 2x+y=0\]
\[\Rightarrow y=-2x\]
Now substituting this value of \[y\] in our original function
\[\Rightarrow {{x}^{2}}+x(-2x)+{{(-2x)}^{2}}=27\]
\[\Rightarrow 3{{x}^{2}}=27\]
\[\Rightarrow x=\pm 3\]
On putting these values in the given function \[{{x}^{2}}+xy+{{y}^{2}}=27\] we get,
\[\begin{align}
  & x=3 \\
 & \Rightarrow y=-6,3 \\
 & x=-3 \\
 & \Rightarrow y=3,-6 \\
\end{align}\]
\[\Rightarrow (3,-6),(3,3),(-3,3),(-3,6)\]
This is not the final coordinates, we have to put this in the \[\dfrac{dy}{dx}\] to check whether it would be zero or not.
On substituting we get,
Only \[(3,-6)\] and \[(-3,6)\] coordinates are satisfying the condition for horizontal tangent.
Now for vertical tangent, \[\dfrac{dy}{dx}\] is not defined
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-(2x+y)}{x+2y}\]
Means for this, \[x+2y=0\]
\[\Rightarrow x=-2y\]
As similar to the above, we get
\[\Rightarrow (-6,3),(3,3),(6,-3),(-3,-3)\]
On putting these coordinates, we get

Only \[(-6,3)\] and \[(6,-3)\] coordinates satisfy for vertical tangent.

Note:
The implicit differentiation means the differentiation of an implicit function that is something variable equal to constant means we can’t get the value of a variable directly by knowing the other while the explicit function is the representation of a variable in terms of others.