Question

The equation of the common tangents to the two hyperbolas \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\) and \(\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1\), are

• A. \(y=\pm x \pm \sqrt{b^2-a^2}\)
• B. \(y=\pm x \pm \sqrt{a^2-b^2}\)
• C. \(y=\pm x \pm \sqrt{a^2+b^2}\)
• D. \(y=\pm x \pm \sqrt{a^2-b^2}\)

Hint:

Common tangents of symmetric hyperbolas often come in pairs \(y=\pm x+c\). Use tangent condition \(y=mx\pm\sqrt{a^2m^2-b^2}\) and set \(m=\pm1\).

Answer 1

The Correct Option is B

Step 1: Assume common tangent form.
Since hyperbolas are symmetric about both axes and interchange \(x,y\), common tangents will be symmetric lines of form:
\[ y=mx+c \] Step 2: Condition for tangency to \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\).
For hyperbola \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\), tangent with slope \(m\) is:
\[ y=mx \pm \sqrt{a^2m^2-b^2} \] Step 3: Condition for tangency to \(\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1\).
Interchanging \(x\) and \(y\) gives tangent:
\[ x=my \pm \sqrt{a^2m^2-b^2} \] Step 4: For same line to be tangent to both.
Common tangents occur when \(m=\pm 1\).
Substituting \(m=1\):
\[ y=x\pm \sqrt{a^2-b^2} \] Substituting \(m=-1\):
\[ y=-x\pm \sqrt{a^2-b^2} \] So combined form:
\[ y=\pm x \pm \sqrt{a^2-b^2} \] Final Answer: \[ \boxed{y=\pm x \pm \sqrt{a^2-b^2}} \]