Question

How do you find all points on the curve ${x^2} + xy + {y^2} = 7$ where the tangent line is parallel to the x-axis, and the point where the tangent line is parallel to the y-axis?

Answer 1

Hint: This problem comes under differentiation of curves. Differentiation is the process of finding the rate of change of function in one variable with respect to another variable. In this problem we need to find the two tangent lines (points) which are parallel to both x and y axis. In general tangent line means a line which passes through the curve at only one point. We solve this by differentiation and by complete step by step explanation.

Complete step-by-step solution:
First we need to find $\dfrac{{dy}}{{dx}}$ for horizontal line for x axis
Now consider the equation ${x^2} + xy + {y^2} = 7$
Differentiate on both sides, we get $x = 1$
$\dfrac{{dy}}{{dx}}({x^2} + xy + {y^2}) = \dfrac{{dy}}{{dx}}(7)$
Now applying differentiation formula, we get
$\Rightarrow$$\dfrac{{dy}}{{dx}}{x^2} + \dfrac{{dy}}{{dx}}xy + \dfrac{{dy}}{{dx}}{y^2} = \dfrac{{dy}}{{dx}}7$
$\Rightarrow$$2x + 1y + x\dfrac{{dy}}{{dx}} + 2y\dfrac{{dy}}{{dx}} = 0$
Now separate $\dfrac{{dy}}{{dx}}$, we get
$\Rightarrow$$\dfrac{{dy}}{{dx}}(x + 2y) = - (2x + y)$
$\Rightarrow$$\dfrac{{dy}}{{dx}} = \dfrac{{ - (2x + y)}}{{(x + 2y)}}$
Now $\dfrac{{dy}}{{dx}} = 0$ when the numerator is zero, provided that this does not make the denominator 0
Therefore $2x + y = 0,y = - 2x$
Now, we have two equations
${x^2} + xy + {y^2} = 7 - - - - - (1)$
$y = - 2x - - - - - - (2)$
Now, substitute equation 2 in 1 by substitution method, we get
$\Rightarrow$${x^2} + x( - 2x) + {( - 2x)^2} = 7$
On squaring the term and we get
$\Rightarrow$${x^2} - 2{x^2} + 4{x^2} = 7$
On adding we get
$\Rightarrow$$5{x^2} - 2{x^2} = 7$
Let us subtract the term and we get,
$\Rightarrow$$3{x^2} = 7$
On divide the term and we get,
$\Rightarrow$${x^2} = \dfrac{7}{3}$
Taking square root on both sides, we get
$\Rightarrow$$x = \pm \sqrt {\dfrac{7}{3}} $
Now, removing root in denominator we get
$\Rightarrow$$x = \pm \sqrt {\dfrac{7}{3}} \times \sqrt {\dfrac{3}{3}} $
On rewriting we get
$\Rightarrow$$x = \pm \sqrt {\dfrac{{7 \times 3}}{{3 \times 3}}} $
On multiply the term and we get
$\Rightarrow$$x = \pm \dfrac{{\sqrt {21} }}{3}$
Now sub $x$ in equation 2, we get
$\Rightarrow$$x = + \dfrac{{\sqrt {21} }}{3},y = - \dfrac{{2\sqrt {21} }}{3}$
$\Rightarrow$$x = - \dfrac{{\sqrt {21} }}{3},y = + \dfrac{{2\sqrt {21} }}{3}$
The tangent to the curve is horizontal at the two points
$\Rightarrow$$\left( {\dfrac{{\sqrt {21} }}{3}, - \dfrac{{2\sqrt {21} }}{3}} \right)\left( { - \dfrac{{\sqrt {21} }}{3}, + \dfrac{{2\sqrt {21} }}{3}} \right)$
Now we find points at which is tangent is vertical,
Now $\dfrac{{dy}}{{dx}} = 0$ when the denominator is zero, provided that this does not make the numerator 0
Similarly,
$x + 2y = 0$
$x = - 2y - - - - - - (3)$
Now substitute equation 2 in 3 by substitution method, similarly we get
$\Rightarrow$$y = \pm \dfrac{{\sqrt {21} }}{3}$
Now substitute $y$ in equation 2, we get
$\Rightarrow$$y = + \dfrac{{\sqrt {21} }}{3},x = - \dfrac{{2\sqrt {21} }}{3}$
$\Rightarrow$$y = - \dfrac{{\sqrt {21} }}{3},x = + \dfrac{{2\sqrt {21} }}{3}$

Therefore the points on the curve at which the tangent is vertical are
$\left( { - \dfrac{{\sqrt {21} }}{3},\dfrac{{2\sqrt {21} }}{3}} \right)\left( {\dfrac{{\sqrt {21} }}{3}, - \dfrac{{2\sqrt {21} }}{3}} \right)$

Note: Differentiation is the rate of change of function in the process of finding derivatives. The tangent is a meeting point of straight line and the curve there will be only one point of contact. We solve this by differentiating the given equation and then by using the x and y axis concept. Then the concept of making equation numerator and denominator of the derivatives we found. And then we obtain two equations; we solve them by substitution method; we find the points of the curve and tangent for both horizontally and vertically.