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 $m_2$




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?

  • $\begingroup$ To find the hyperbola, you need to find $a, b.$ $\endgroup$
    – Allawonder
    Apr 18 '20 at 19:59
  • $\begingroup$ 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 $b^2$ $\endgroup$
    – user744725
    Apr 18 '20 at 20:02
  • $\begingroup$ 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) $\endgroup$
    – Tavish
    Apr 18 '20 at 20:08
  • $\begingroup$ @Tavish The method you propose for the second equation doesn’t generate an indepenent equation. The two constraints are equivalent. $\endgroup$
    – amd
    Apr 18 '20 at 20:48
  • $\begingroup$ @amd Hmm, I see. $\endgroup$
    – Tavish
    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:

hyperbola, tangents and intercept circle

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


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



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.


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