Question

$ABC$ is an isosceles triangle with its base $BC$ twice its altitude. A point $P$ moves within the triangle such that the square of its distance from $BC$ is half the area of rectangle contained by its distances from the two sides . The locus of $P$ is an ellipse with eccentricity is

• A. $\sqrt{\frac{2}{3}}$
• B. $\sqrt{\frac{3}{2}}$
• C. $\sqrt{2\frac{2}{3}}$
• D. $\sqrt{2\frac{3}{2}}$

Answer 1

The correct option is A $\sqrt{\frac{2}{3}}$

Let $AF$ be the perpendicular from $A$ to $BC$ as shown in the figure. Then according to the given data and the property of isosceles triangles, we have:

$AF=BF=FC\Rightarrow \angle BAC={90}^o,\angle ABC=\angle ACB={45}^o$

So, we let $BC$ be the $x-axis$ and points $A,B,C$ be as shown in the figure. Point $P=(x,y)$ is an arbitrary point inside triangle $ABC$.

Thus, equation of line $AB$ is $x-y=0$

and equation of line $AC$ is $x+y-2=0$

We are given that:

$(PF{)}^2=\frac{PD.PE}{2}$

Note - it is important to consider the modulus in the distance formula of a point from a line, so with proper sign, we get:

$\Rightarrow y^2=\frac{(y-x)(x+y-2)}{2\sqrt{2}.\sqrt{2}}$

$\Rightarrow 4y^2=y^2-x^2-2(y-x)$

$\Rightarrow x^2+3y^2+2y-2x=0$

$\Rightarrow (x-1{)}^2+3(y+\frac{1}{3}{)}^2=\frac{4}{3}$

This gives the eccentricity as $\sqrt{1-\frac{1}{3}}=\sqrt{\frac{2}{3}}$