Question

Let $A$ and $B$ be two drawing pins such that $AB=10$ units .Consider a string whose end points are at $A,B$ and length is of $16$ units. If the point of pencil moves on paper such that fixed ends of string are always tight the area encolsed by curve traced by pencil is

• A. $36\pi$ sq. units
• B. $100\pi$ sq. units
• C. $16\sqrt{39}\pi$ sq. units
• D. $8\sqrt{39}\pi$ sq. units

Answer 1

The correct option is D $8\sqrt{39}\pi$ sq. units

Locus of a point whose sum of distances from two fixed points(foci) is constant is an ellipse.
Location of pins (fixed points) are at $A$ and $B$
and let location of pencil at any time be $P$
So, $AP+BP=2a$
$\Rightarrow 2a=16\Rightarrow a=8$
distance between pins is $10$
$\Rightarrow 2ae=10\Rightarrow e=\frac{10}{16}=\frac{5}{8},[\because a=8]$
$\because e^2=1-\frac{b^2}{a^2}$
$\Rightarrow \frac{b^2}{a^2}=1-e^2$
$\Rightarrow \frac{b^2}{64}=1-\frac{25}{64}$
$\Rightarrow b^2=39\Rightarrow b=\sqrt{39}$
Area of an ellipse $=\pi ab$
Area $=\pi \times 8\times \sqrt{39}$
$=8\sqrt{39}\pi$ sq. units