Find points of the parallelogram which sides are tangent to the hyperbola

Question

Find points of the parallelogram which sides are tangent to the hyperbola, if the hyperbola’s equation is: $ 2x^2−4xy+y^2−2x+6y−3=0$ and one of the points is $P_{1}(3,4)$.

I’ve tried to find tangents to the hyperbola, which contain point $P_{1}$, but it get’s too messy.

Answer 1

indicates to me that the left and top sides of the parallelogram can be tangent to the parallelogram, but the right and bottom might not, looking at the top right. Or, maybe the top of the parallelogram goes down and to the left. Either way, for this hyperbola, $\frac {dy} {dx} is:$

$$\frac {dy} {dx} = \frac {(-2 x + 2 y + 1)} {(-2 x + y + 3)}$$ And the formula of any line created by a point on the parabola and (3,4) is: $$y − y_1 = \frac {y_2 - y_1} {x_2 - x_1} (x − x_1)$$ $$\implies y − y_1 = \frac {4 - y_1} {3 - x_1} (x − x_1)$$ So the slope of any line created by a point on the parabola and (3,4) is: $$ \frac {4 - y_1} {3 - x_1}$$ And where $\frac {dy} {dx} =\frac {4 - y_1} {3 - x_1}$, the slopes are equivalent.

And where $ 2x^2−4xy+y^2−2x+6y−3=0$, valid tangent points of the sides of the parallelogram exist.

Therefore, each tangent point (TP) on the parallelogram must satisfy the following equations: $$ 2x^2−4xy+y^2−2x+6y−3=0$$ $$\frac {4 - y} {3 - x} = \frac {(-2 x + 2 y + 1)} {(-2 x + y + 3)}$$ This gives 2 equations 2 unknowns, which can be solved: (link to sys of equations on wolframalpha.com)

There is just 1 solution point given by this equation: (1,-3).

But, if you take a look at the graph of $\frac {4 - y} {3 - x} = \frac {(-2 x + 2 y + 1)} {(-2 x + y + 3)}$, you'll find that there is a hole in the plot, where actually both sides of the equation are undefined: $\frac {4 - y} {3 - x} = \frac {(-2 x + 2 y + 1)} {(-2 x + y + 3)}$" /> In that case, $(-2 x + y + 3)=0$ and $(3-x)=0$, or $y=2x-3$ and $x=3$ (i.e. (3,3)). Does this solution satisfy the equation of the hyperbola as well? $$ 2(3)^2−4((3)(3))+(3)^2−2(3)+6(3)−3 \space ^{?} _{=} \space 0$$ Yes.

So, (3,3) is a tangent point as well.

Given that $$ (3,4), (3,3), (1,-3)$$

are tangent points, the parallelogram can be constructed. $$ (3,4), (3,3), (1,-3), (1,-2)$$

EDIT: One thing that may help a future viewer of this answer is the parameterization of the hyperbola:

the parametric form for the hyperbola, $$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$$ is $$(h+a\sec \theta, k+b\tan \theta).$$

Rearranging the given equation: $$2x^2−4xy+y^2−2x+6y−3=0$$ $$2x^2−4xy+y^2−2x+6y=3$$ $$y (-4 x + y + 6) + x (2 x - 2)=3$$