In triangle ABC, $a=4,b=c=2\sqrt{2}$. A point $P$ moves in the triangle such that the square of it's distance from $BC$ is half the area rectangle contained by its distances from the other two sides.if $D$ be the center of locus of $P$. Then,Find Locus of $P$, it's eccentricity and Area of Quadrilateral $ABCD$
My attempt:
Let $A$ be the origin. Since $ABC$ is a right=angled triangle, let the other two sides be $X$-axis,$Y$-axis, with $B$ on $Y-axis$. Let the point $P=(x,y)$
so $A=(0,0),B=(0,2\sqrt{2}),C=(2\sqrt{2},0)$
distance from line $BC$ is $\frac{|x+y-2\sqrt{2}|}{\sqrt{2}}$
area of the rectangle formed by $P$ with other two sides is $xy$ so
$$\implies \frac{xy}{2}=\left(\frac{|x+y-2\sqrt{2}|}{\sqrt{2}}\right)^2$$
$$\implies xy=x^2+y^2+8+2xy-4\sqrt{2}x-4\sqrt{2}y$$
$$\implies x^2+y^2+8+xy-4\sqrt{2}x-4\sqrt{2}y=0$$
I know that this is an ellipse but couldn't find a way to simplify it further to find the eccentricity in a simple way.
since the equation of ellipse is symmetric about $x=y$,center of ellipse lies on it ,so
solving $x=y$ and the given ellipse we get,
$$3x^2-8\sqrt{2}x+8=0$$ as the center is mid-point of the solutions of $x=y$ and the ellipse we have center $D$ as, $$D=\left(\frac{4\sqrt{2}}{3},\frac{4\sqrt{2}}{3}\right)$$ after getting the center and one of the axis,I can now find the eccentricity of the ellipse,but It has a lot of calculation part and looks somewhat messy.
I also know that I can solve the equation by rotating the ellipse.
Now,my question is Is there any analytical method to say that locus is ellipse,without actually writing the equations?
Is there a way I can convert equation of ellipse to the form $\frac{SP}{PM}=e$?( $S$ is focus,$M$ is directrix)?
Is there a way to find the eccentricity of ellipse in general form without rotating it,and converting it to the form in the above question ( I find converting to the form $\frac{SP}{PM}=e$ pretty hard because it involves manipulating,so I don't want to prefer this method but I would like to see the equation converted)
Thank you.
You can avoid the rotated ellipse if you instead place the triangle on the coordinate system at $A = (0,0)$, $B = (2,2)$, $C = (2,-2)$. Then the area of the rectangle with opposite vertices at $A$ and $P = (x,y)$ is $$\frac{1}{2}|x^2 - y^2|.$$ Since $P$ is located inside the triangle, we must have $|y| \le x$, so we can remove the absolute value signs. Therefore, the locus of $P$ satisfies $$\frac{1}{4}(x^2 - y^2) = (x-2)^2.$$ This very easily leads to the desired ellipse.
As for the motivation to place the triangle in this way, it should be clear that the locus of $P$ is necessarily symmetric about the altitude from $A$ to $BC$, because the triangle itself is symmetric about this altitude. So whatever curve it is, if you put the triangle on the coordinate plane in such a way that this altitude coincides with a coordinate axis, the resulting locus will respect that symmetry.