Determining the value of a constant in a multicalc exercise regarding a tangent and a curve

Question

Let the line that goes through the points $(1,0)$ and $(-1,2)$ be a tangent to the curve defined by $x^2-2y^2=C\in\mathbb{R}$. I need to determine the value of $C$. My strategy thus far has been to find the point of intersection $(a,b)$ and then use that to determine the value of $C$. Thus far I have that $ax-2by=C$ and $a+2b=0$. I don't really know what to do here, some help would be greatly appreciated.

Answer 1

You don't need derivatives. Note that the equation for your line is $$y=1-x$$ So you need to find the intersection point. When replacing this into the conic equation, you have $$x^2-2(1-x)^2= C$$ Therefore $$x^2-4x+2+C=0$$ You can now solve for the tangent points and draw conclusions.

Answer 2

If you don’t happen to recognize that the curve is a conic section, allowing you to solve this by examining the discriminant of a quadratic, you can proceed more generically as follows: If a line is tangent to a curve, it is orthogonal to the curve’s normal at the point of tangency. The direction of the normal to this curve is given by the gradient $\nabla(x^2-2y^2) = (2x,-4y)$, the direction vector of the line is $(1,0)-(-1,2)=(2,-2)$ and the orthogonality condition can be expressed via a dot product: $$(2x,-4y)\cdot(2,-2)=4x+8y=0.$$ The intersection of this line with the given one is the point of tangency to the curve, and $C$ is then found by plugging the coordinates of this point into $x^2-2y^2$.