Question

Find the equation, the length, and the common tangent to the two hyperbolas $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$.

Answer 1

Hint: Use the condition that a line y = mx + c is a tangent to a hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ if ${{a}^{2}}{{m}^{2}}-{{b}^{2}}={{c}^{2}}$ and the line y = mx + c is a tangent to the hyperbola $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$ if ${{a}^{2}}-{{b}^{2}}{{m}^{2}}={{c}^{2}}$.

Complete step by step answer:
Before proceeding with the question, we must know the formula like distance formula, condition of tangency to a hyperbola.
If we have given a hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, then, a and b are the respectively the major and minor axis of the hyperbola. In this question, we have been given that the two hyperbolas $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$. We have to find the length and equation of the common tangent.
Let us assume y to be the common tangent to the given hyperbolas, y = mx + c.
We know that condition for y = mx + c to be the tangent of the hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is ${{a}^{2}}{{m}^{2}}-{{b}^{2}}={{c}^{2}}$ and for hyperbola $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$, it is ${{a}^{2}}-{{b}^{2}}{{m}^{2}}={{c}^{2}}$.
So, we can number the equations as:
For hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, ${{a}^{2}}{{m}^{2}}-{{b}^{2}}={{c}^{2}}.....(i)$
For hyperbola $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$, ${{a}^{2}}-{{b}^{2}}{{m}^{2}}={{c}^{2}}.....(ii)$
Equating equations (i) and (ii) we get,
${{a}^{2}}{{m}^{2}}-{{b}^{2}}={{a}^{2}}-{{b}^{2}}{{m}^{2}}$
$\Rightarrow {{m}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right)={{a}^{2}}+{{b}^{2}}$
Cancelling out the similar terms we get,
$\Rightarrow {{m}^{2}}=1$
$\therefore m=\pm 1$
Putting the value of ${{m}^{2}}$ in equation (i) we get,
${{c}^{2}}={{a}^{2}}-{{b}^{2}}$
$\therefore c=\pm \sqrt{{{a}^{2}}-{{b}^{2}}}$
Putting the values of $m$ and $c$ in the equation of the line y = mx + c, we get the equation of tangent as,
$y=\pm x\pm \sqrt{{{a}^{2}}-{{b}^{2}}}$
We can write the equation of tangent by dividing both sides by $\sqrt{{{a}^{2}}-{{b}^{2}}}$ as $\dfrac{y}{\sqrt{{{a}^{2}}-{{b}^{2}}}}=\pm \dfrac{x}{\sqrt{{{a}^{2}}-{{b}^{2}}}}+1$
Collecting the variables on left-hand-side we get,
$\dfrac{y}{\sqrt{{{a}^{2}}-{{b}^{2}}}}\mp \dfrac{x}{\sqrt{{{a}^{2}}-{{b}^{2}}}}=1.....(iii)$
We know that the tangent to the hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ at $\left( {{x}_{1}},{{y}_{1}} \right)$ is given by $\dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1$.
Therefore, on comparing equation (iii) with $\dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1$ we get the value of ${{x}_{1}}=\pm \dfrac{{{a}^{2}}}{\sqrt{{{a}^{2}}-{{b}^{2}}}}$ and${{y}_{1}}=\pm \dfrac{{{b}^{2}}}{\sqrt{{{a}^{2}}-{{b}^{2}}}}$.
Similarly, on comparing for hyperbola $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$ we get, ${{x}_{2}}=\pm \dfrac{{{b}^{2}}}{\sqrt{{{a}^{2}}-{{b}^{2}}}},{{y}_{2}}=\pm \dfrac{{{a}^{2}}}{\sqrt{{{a}^{2}}-{{b}^{2}}}}$.
Now, we have already found out points $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$. Therefore, the length of common tangent can be found by using the distance formula given by
$\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
$\Rightarrow \sqrt{{{\left( \pm \dfrac{{{b}^{2}}}{\sqrt{{{a}^{2}}-{{b}^{2}}}}-\left( \pm \dfrac{{{a}^{2}}}{\sqrt{{{a}^{2}}-{{b}^{2}}}} \right) \right)}^{2}}+\left( \pm \dfrac{{{a}^{2}}}{\sqrt{{{a}^{2}}-{{b}^{2}}}}-\left( \pm \dfrac{{{b}^{2}}}{\sqrt{{{a}^{2}}-{{b}^{2}}}} \right) \right)}$
$\Rightarrow \sqrt{\left( \dfrac{{{b}^{4}}}{{{a}^{2}}-{{b}^{2}}}+\left( \dfrac{{{a}^{4}}}{{{a}^{2}}-{{b}^{2}}} \right)+\dfrac{2{{a}^{2}}{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}} \right)+\left( \dfrac{{{a}^{4}}}{{{a}^{2}}-{{b}^{2}}}+\left( \dfrac{{{b}^{4}}}{{{a}^{2}}-{{b}^{2}}} \right)+\dfrac{2{{a}^{2}}{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}} \right)}$
Collecting the same terms together we get:
$\Rightarrow \sqrt{\left( \dfrac{2{{b}^{4}}}{{{a}^{2}}-{{b}^{2}}}+\left( \dfrac{2{{a}^{4}}}{{{a}^{2}}-{{b}^{2}}} \right)+\dfrac{4{{a}^{2}}{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}} \right)}$
Taking out 2 as common we get:
$\Rightarrow \sqrt{\dfrac{2\left( {{a}^{4}}+{{b}^{4}}+2{{a}^{2}}{{b}^{2}} \right)}{{{a}^{2}}-{{b}^{2}}}}$
$\Rightarrow \sqrt{\dfrac{2{{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}}{{{a}^{2}}-{{b}^{2}}}}$
Taking out square out of the root we get:
$\Rightarrow \left( {{a}^{2}}+{{b}^{2}} \right)\sqrt{\dfrac{2}{{{a}^{2}}-{{b}^{2}}}}$
Therefore, after simplification, we get the length of the common tangent is equal to, $\left( {{a}^{2}}+{{b}^{2}} \right)\sqrt{\dfrac{2}{{{a}^{2}}-{{b}^{2}}}}$.
And the equation of common tangent we have already found out as, $y=\pm x\pm \sqrt{{{a}^{2}}-{{b}^{2}}}$. Hence, the equation of common tangent is $y=\pm x\pm \sqrt{{{a}^{2}}-{{b}^{2}}}$ and length of the common tangent is $\left( {{a}^{2}}+{{b}^{2}} \right)\sqrt{\dfrac{2}{{{a}^{2}}-{{b}^{2}}}}$.

Note: We must be very careful about the values of m. After taking the root, we must not forget to take the negative values also. The possibility of mistake is by adding the coordinates in the distance formula instead of subtracting the coordinates and ending up with the wrong distance.