# Pair of tangents from point $(2\sqrt2,1)$ to hyperbola $\frac{x^2}{a^2} -\frac{y^2}{b^2} = 1$

From a point $$(2\sqrt2,1)$$ a pair of tangents are drawn to $$\frac{x^2}{a^2} -\frac{y^2}{b^2} = 1$$ which intersect the coordinate axes in concyclic points. If one of the tangents is inclined at an angle of $$\arctan\frac{1}{\sqrt{2}}$$ with the transverse axis of the hyperbola, then find the equation of the hyperbola and also the circle formed using the concyclic points.

My Attempt

A tangent to the hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$ with slope $$m$$ is given by $$y=mx±\sqrt{a^2m^2-b^2}$$ Plugging $$(2\sqrt2,1)$$ in this equation, I get $$m^2(8-a^2)+m(-4\sqrt2)+(1+b^2)=0$$ This equation gives two values of $$m$$

$$m_1=\frac{1}{\sqrt2}$$ and $$m_2$$

$$m_1+m_2=\Large\frac{4\sqrt2}{8-a^2}$$

And

$$m_1m_2=\Large\frac{1+b^2}{8-a^2}$$

How do I proceed further? I know we have to use the fact that the points at which the tangents intersect the axes are concyclic. How do I apply this and get the required result or is there another easy way to do this?

• To find the hyperbola, you need to find $a, b.$ Apr 18 '20 at 19:59
• Yes I know that I was thinking if I can find $m_2$ then I can solve for $a^2$ and then use it get the value of $b^2$ Apr 18 '20 at 20:02
• Plug in $m=m_1, x=2\sqrt 2, y=1$ in the equation of the tangent with the minus sign to get one equation in $a,b$. For the second equation, you have to solve for the intersection point of this tangent (with slope $\frac{1}{\sqrt 2}$ ) , find the slope of the hyperbola at that point using differentiation and equate it to $\frac{1}{\sqrt 2}$ . Then you can solve for $a$ and $b$. (Good luck) Apr 18 '20 at 20:08
• @Tavish The method you propose for the second equation doesn’t generate an indepenent equation. The two constraints are equivalent.
– amd
Apr 18 '20 at 20:48
• @amd Hmm, I see. Apr 18 '20 at 20:53

You’re given the slope of one of the tangent lines, so you’ve got its equation: $$x-y\sqrt2=\sqrt2.\tag1$$ Its axis intercepts are $$\sqrt2$$ and $$-1$$, respectively. The one-parameter family of circles that pass through these two points have equations $$x(x-\sqrt2)+y(y+1) + \lambda(x-y\sqrt2-\sqrt2)=0.\tag2$$ The first two terms represent a circle with diameter given by the above intercepts. The other two intersections of this circle with the coordinate axes work out to be $$(-\lambda,0)$$ and $$(0,\lambda\sqrt2)$$. The line through these two points has an equation of the form $$x\sqrt2-y+\lambda\sqrt2=0.\tag3$$ Even without knowing $$\lambda$$, you can extract its slope. Plugging the two known slopes into your generic equation of the tangent generates a system of two equations in $$a$$ and $$b$$ that you can solve.

To find the circle through the intercepts, you can substitute $$x=2\sqrt2$$ and $$y=1$$ into equation (3), solve for $$\lambda$$, and plug that into equation (2).

The end result is illustrated below:

N.B.: This solution assumes that there are four axis intercepts. There’s another solution with a vertical tangent, so that there are only three intercepts.

Given the point $$(2\sqrt2,1)$$ and the slopes $$\frac1{\sqrt2}$$, $$m$$, the equations of the two tangent lines are $$y-1 =\frac1{\sqrt2}( x-2\sqrt2), \>\>\>\>\>y-1 = m(x-2\sqrt2)$$

which intersect the axes at $$A(\sqrt2,0)$$, $$B(0,-1)$$ and $$C(2\sqrt2-\frac1m, 0)$$, $$D(0, 1-2\sqrt2m)$$, respectively. Given that $$A$$, $$B$$, $$C$$ abd $$D$$ are concyclic, we have $$\angle ACB = \angle ADB=\theta$$, i.e.

$$\tan\theta=\frac {BO}{CO}=\frac {AO}{DO} \implies \frac1{2\sqrt2-\frac1m}=\frac{\sqrt2}{2\sqrt2m-1}$$

which leads to $$m=\sqrt2$$. The tangent line equations to the hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$ is given by $$(y-m x)^2=a^2m^2-b^2$$

Substitute the point $$(2\sqrt2,1)$$ and the slopes $$m=\frac1{\sqrt2},\>\sqrt2$$ into the equations to get

$$2a^2-b^2=9,\>\>\>\>\>\frac12a^2 -b^2 = 1$$

Solve to obtain $$a^2=\frac{16}{3}$$, $$b^2=\frac53$$ and the equation of the hyperbola

$$\frac{3x^2}{16}-\frac{3y^2}{5}=1$$

From the known axis intersections, the cyclic circle is obtain as,

$$(x-\frac5{2\sqrt2})^2+(y+2)^2=\frac{33}8$$

Tangents can intersect the axes at four concyclic points (thanks to amd for correcting me), but they could also intersect at only THREE points: this is not explicitly ruled out by the text of the problem.

As no tangent can pass through $$(0,0)$$, this is only possible if a tangent is parallel to $$y$$-axis, i.e. it has equation $$x=2\sqrt2$$. Hence $$a=2\sqrt2$$ and the equations of both tangents are known. one can then easily find that $$b=\sqrt3$$ all the other requested things.