Hyperbola
Circle tangent to asymptotes and tangent lines — pentagon area
MJAT_TS5_P1
Grade 12
Question:
Two tangents, one from $A(2,1)$ and other from $B(-2,1)$, are drawn to the hyperbola $\dfrac{x^2}{4}-y^2=1$. A circle which touches these two tangents and two asymptotes of the hyperbola has its centre at $(a,\lambda)$ where $\lambda>0$. The least area of the pentagon in which this circle is inscribed (two sides are asymptotes, two sides are the tangents) is:
A) $(3+\sqrt{5})$ sq.units
B) $(3+2\sqrt{5})$ sq.units
C) $(6+2\sqrt{5})$ sq.units
D) $(6+4\sqrt{5})$ sq.units
Step-by-Step Solution
Key Concept: Asymptotes of $x^2/4-y^2=1$: $y=\pm x/2$. Tangent from $A(2,1)$: substitute and find tangent line. Tangent from $B(-2,1)$: by symmetry. The circle touches both asymptotes and both tangents — find centre and radius, then compute pentagon area.
Least area $=\mathbf{(6+4\sqrt{5})}$ sq.units. Answer: D.
Correct Answer: D