Draw the tangent to the hyperbola with equation $x^2-3y^2=36$ through the point $A(0;-2),$ as $A$ is not from the hyperbole

Question

Draw the tangent to the hyperbola with equation $$x^2-3y^2=36$$ through the point $A(0;-2),$ as $A$ is not from the hyperbole.

I don't know how to approach problems like this one. Is there a general approach or it depends on each problem?

I have seen a method for solving such problems and it basically says that we should write the equation of the hyperbole as a function, so in our case we have $$3y^2=x^2-36\Rightarrow y^2=\dfrac{x^2-36}{3}\Rightarrow y=\pm\sqrt{\dfrac{x^2-36}{3}}$$ I am not sure how this helps though and how we are supposed to use it to solve our problem.

I have also calculated the first derivative of the function (if it somehow helps) $y,$ so $$y'=\pm\dfrac{x}{3\sqrt{\dfrac{x^2-36}{3}}}$$

Can you give me some general approaches for problems like this (tangents to hyperbolas, parabolas, ellipses through points lying on them, and also through external points)?

Answer 1

I actually haven't learnt hyperbola yet, but as you seek for a general approach (this works good in circles, parabolas and ellipses, maybe also in hyperbola but I don't know), I have some for you.

If a point $(x_1,y_1)$ is outside the curve (two tangents can be drawn to the curve) then the equation of the pair of tangents is $$SS_1=T^2$$

If a point $(x_1,y_1)$ is on the curve (this theory holds good for hyperbolas too) then equation of tangent is $$T=0$$

Let the equation of curve be $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ then $$S=ax^2+2hxy+by^2+2gx+2fy+c$$ $$S_1=a{x_1}^2+2hx_1y_1+b{y_1}^2+2gx_1+2fy_1+c$$ $$T=axx_1+byy_1+g(x+x_1)+f(y+y_1)+h(xy_1+yx_1)+c$$

Answer 2

Alternative approach:

Often, in a situation like this, the problem composer intends that you look for a way to prevent the Math from being ugly.

In order for a point $~(x,y)~$ to represent the point (or one of the points) on the hyperbola such that the line segment $~\overline{(x,y) ~(0,-2)}~$ is tangent to the hyperbola, two constraints have to be satisfied:

• Point $~(x,y)~$ must be on the hyperbola.
  This implies that $~x^2 - 3y^2 = 36.$
  Note that $~y = 0 \implies x = \pm 6.$

• The slope of the line $~\overline{(x,y) ~(0,-2)}~$ must be identical to the slope of the hyperbola at the point $~(x,y).~$
  Using implicit differentiation, you have that
  $2x - 6yy' = 0.~$
  Note that $~y = 0 \implies x = 0.~$
  This conflicts with the conclusion in the first bullet point above.
  Therefore, you can assume, without loss of generality, that $~y \neq 0.$
  Therefore, you are allowed to divide the equation by $~y~$ to conclude that $~y' = \dfrac{x}{3y}.$

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The second bullet point above implies that

$$\frac{y - (-2)}{x - (0)} = y' = \frac{x}{3y} \implies $$

$$\frac{y+2}{x} = \frac{x}{3y} \implies $$

$$3y^2 + 6y = x^2 \implies $$

$$x^2 - 3y^2 = 6y. \tag1 $$

Compare the constraint in (1) above to the constraint within the first bullet point, at the start of my answer.

At this point, you still have not determined whether one or more tangent lines are feasible. However, what you have determined is that if one or more tangent lines are feasible, each such corresponding point must satisfy:

$$36 = x^2 - 3y^2 = 6y \implies 36 = 6y \implies y = 6. \tag2 $$

So, what you now know is that it is impossible to construct the corresponding tangent lines if the value of $~y~$ is not equal to $~6.~$

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Personally, I refer to $~y = 6~$ as a candidate value.

So, I will complete the problem by manually checking which candidate values of $~x~$ correspond to the candidate value of $~y = 6,~$ and then I will manually check whether the corresponding line segment $~\overline{(x,6) ~(0,-2)}~$ is in fact tangent to the hyperbola at the point $~(x,6).$

$$y = 6 \implies x^2 = 36 + [ ~3 \times (36) ~] = 144 = 12^2.$$

Therefore, the two candidate points to examine are :

• $(x,y) = (12,6) \implies y' = \dfrac{x}{3y} = \dfrac{2}{3}.$
  
  Then the slope of the line $~\overline{(12,6) ~(0,-2)}~$ is $~\dfrac{8}{12} = \dfrac{2}{3}.~$
  
  The candidate value of $~(12,6)~$ is accepted.
  
  

• $(x,y) = (-12,6) \implies y' = \dfrac{x}{3y} = \dfrac{-2}{3}.$
  
  Then the slope of the line $~\overline{(-12,6) ~(0,-2)}~$ is $~\dfrac{8}{-12} = \dfrac{-2}{3}.~$
  
  The candidate value of $~(-12,6)~$ is accepted.

Therefore, the two tangent lines are
$\overline{(12,6) ~(0,-2)}~$ and $~\overline{(-12,6) ~(0,-2)}.$

Answer 3

Yet another solution: the polar of $A$ is $0\cdot x-3\cdot(-2)\cdot y = 36$, that is $y=6$ from where the points of tangency, namely $(\pm12,6)$ are easily calculated.