Ellipse, circle and square.

Geometry Level 3

An ellipse $E_1$ has the following properties:-

The major axis is parallel to x-axis.
Its center $F(a,b)$ is such that $a < 0$ and $b > 0$ .
The co-ordinate axes are tangents to it.
The semi-major and semi-minor lengths have unequal integral values.

Now, an ellipse $E_2$ is drawn by reflecting $E_1$ about the line $y = x$ .

The co-ordinate axes touch $E_1$ at $A , B$ and $E_2$ at $C , D$ . Normals are drawn to $E_1$ at $A$ and $B$ . Similarly, normals are drawn to $E_2$ at $C$ and $D$ .

The square formed by the point of intersection of these normals has an area of $16$ sq. units.

The equation of the circle circumscribing this square has the following equation:- $x^2 \ + \ y^2 \ + \ 2g x \ + \ 2f y + \ c \ = \ 0$ .

Submit the value of $g \ + \ f \ + \ c$ .

The answer is -4.

1 solution

Ashish Menon
May 28, 2018

Let the equation of $E_1$ be $\dfrac{(x - a)^2}{a^2} \ + \ \dfrac{(y - b)^2}{b^2} \ = \ 1$ .
So, equation of $E_2$ is $\dfrac{(x-b)^2}{b^2} \ + \ \dfrac{(y-a)^2}{a^2} \ = \ 1$ .

From the figure provided above, we deduce that the required normals are $x = a$ , $y = a$ , $x = b$ and $y= b$ .

The area of the square required is $(|a| + b)^2 \ = \ 16$ . Now, since $|a| \neq b$ and $|a|,b \in \mathbb{Z}$ , $|a| = 3 ; b = 1$ .

Furthermore, the center of the required circle is $(-1,-1)$ and the radius is $\sqrt{8}$ units. (see figure).
Hence, the equation of the required circle is $(x+1)^2 \ + \ (y +1)^2 \ = \ 8$ . i.e. $x^2 \ + \ y^2 \ + \ 2x \ + \ 2y \ -6 \ = \ 0$ .

$\therefore \ g \ + \ f \ + \ c \ = \ \color{#3D99F6}{\boxed{-4}}$

Ellipse, circle and square.

The answer is -4.

1 solution

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