Question

How do you find the coordinates of each point on the curve where the tangent line is vertical? The curve is given by ${{x}^{2}}+3{{y}^{2}}=1+3xy$ and $\dfrac{dy}{dx}=\dfrac{3y-2x}{6y-3x}$.

Answer 1

We first find the slope for a vertical line and the slope about the curve. We equate the values in the form of $\dfrac{dy}{dx}=\tan \alpha =\dfrac{3y-2x}{6y-3x}$. The assumed $\left( h,k \right)$ point goes through both the curves and the slope. We solve the equations and find the possible points.

Complete step by step answer:
Let us assume that the coordinates of the point on the curve is $\left( h,k \right)$.
The point $\left( h,k \right)$ is on the curve ${{x}^{2}}+3{{y}^{2}}=1+3xy$.
So, we get ${{h}^{2}}+3{{k}^{2}}=1+3hk$.
We also have that at that point $\left( h,k \right)$, the tangent line is vertical.
All the vertical lines can be represented by $x=K,K\in \mathbb{R}$. The slope of the vertical lines is $\dfrac{\pi }{2}$.
We also know that the slope of any curve at a certain point can be expressed as $\dfrac{dy}{dx}=\tan \alpha$ where the function is $y=f\left( x \right)$ and the angle of the tangent line is $\alpha$.
The particular slope value at point $\left( h,k \right)$ will be ${{\left[ \dfrac{dy}{dx} \right]}_{\left( h,k \right)}}$.
For the curve the slope is $\dfrac{dy}{dx}=\tan \alpha =\dfrac{3y-2x}{6y-3x}$. Now we find the slope at point $\left( h,k \right)$ whose angle is $\dfrac{\pi }{2}$ as the line is vertical.

& \tan \left( \dfrac{\pi }{2} \right)=\dfrac{3k-2h}{6k-3h} \\\ & \Rightarrow 6k-3h=0 \\\ \end{aligned}$$ Simplifying we get $$2k=h$$. We put this value of $$2k=h$$ in the equation of ${{h}^{2}}+3{{k}^{2}}=1+3hk$. ${{\left( 2k \right)}^{2}}+3{{k}^{2}}=1+3\times 2k\times k$. Simplifying we get ${{k}^{2}}=1$ which gives $k=\pm 1$. Putting the value in the equation $$2k=h$$, we get $h=\pm 2$. **Therefore, the possible points are $\left( 2,1 \right)$ and $\left( -2,-1 \right)$.** **Note:** We need to remember the trigonometric value for $$\tan \left( \dfrac{\pi }{2} \right)$$ is undefined as the value tends to infinity as we go close to $$\dfrac{\pi }{2}$$. Therefore, we can represent it as $$\dfrac{1}{0}$$ just to solve the equation, but it is not mathematically correct.