The envelope of line family

Consider a $x''y''$ -coordinate system with origin at the center of the ellipse, and $x''$ and $y''$ axes being along the semi-major and semi-minor axes respectively. Then the equation of the ellipse is $\dfrac {x''^2}{a^2} + \dfrac {y''^2}{b^2} = 1$ , where $a$ and $b$ are the semi-major and semi-minor axes respectively. Rotate the axes by an angle $\theta$ counterclockwise about the origin, the equation in the new $x'y'$ -coordinate system is

$\frac {(x'\cos \theta + y'\sin \theta)^2}{a^2} + \frac {(-x'\sin \theta + y'\cos \theta)^2}{b^2} = 1$

Shifting the origin to $O(h,k)$ , the center of ellipse in the final $xy$ -coordinate system, the equation of the ellipse becomes

$\frac {((x-k)\cos \theta + (y-k)\sin \theta)^2}{a^2} + \frac {(-(x-h)\sin \theta + (y-k)\cos \theta)^2}{b^2} = 1$

We know that $(h,k) = \left(1, \frac 12\right)$ , the midpoint of $AP$ , $\cos \theta = \frac 2{\sqrt 5}$ , and $\sin \theta = \frac 1{\sqrt 5}$ . To find $a$ , we note that vertex of the ellipse is "plotted" by the straight line perpendicular to $AP$ , that is when $A$ , $P$ , and $X$ are colinear and the vertex is the midpoint of $XP$ or $a = \frac {3-AP}2 + OP = \frac {3-\sqrt 5}2-\frac {\sqrt 5}2 = \frac 32$ . The co-vertex is plotted when $XP$ is perpendicular to $AP$ and $b = \frac {XP}2 = \frac {\sqrt{3^2-AP^2}}2 = 1$ . Then the equation:

$\begin{aligned} \frac {4\left((x-1)\frac 2{\sqrt 5} + \left(y-\frac 12 \right)\frac 1{\sqrt 5}\right)^2}9 +\left(-(x-1)\frac 1{\sqrt 5} + \left(y-\frac 12 \right)\frac 2{\sqrt 5}\right)^2 & = 1 \\ \frac {(4x+2y-5)^2}{45} + \frac {(-x+2y)^2}5 & = 1 \\ 16x^2 + 4y^2 + 16xy - 40x - 20y + 25 + 9(x^2 +4y^2-4xy) & = 45 \\ 25x^2 + 40y^2 - 20xy - 40x - 20y + 25 & = 45 \\ 5x^2 + 8y^2 - 4xy -8x-4-4 & = 0 \end{aligned}$

Therefore $A+B+C+D+E+F = 5+8-4-8-4-4 = \boxed{-7}$ .

The family of all these straight lines has equation ( $t$ - parametr) $\\$ $F(x,y,t)=(x-x_0)(P_x-R\cos{t})+(y-y_0)(P_y-R\sin{t})=0$ $\\$
$x_0=\frac{P_x+R\cos{t}}{2}$ $\\$ $y_0=\frac{P_y+R\sin{t}}{2}$ $\\$

The equation for envelope of the family is find from system (eliminate $t$ ) https://brilliant.org/wiki/envelope/

$F(x,y,t)=0$ $\\$ $\frac{\partial F(x,y,t)}{\partial t}=0$ $\\$

Next equation find $\\$ $x P_x+y P_y+\frac{R^2-{P_x}^2-{P_y}^2}{2}=R \sqrt{x^2+y^2}$ $\\$ For $R=3, P_x=2, P_y=1$ squaring the equation and have the ellipse equation

$\\$ $5x^2+8y^2-4xy-8x-4y-4=0.$ $\\$

The answer $\fbox {-7}$ . ​

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Since $X$ is on the circle with equation $x^2 + y^2 = 9$ , let its coordinates be $(p, \sqrt{9 - p^2})$ .

Then the midpoint of $XP$ is $(\frac{p + 2}{2}, \frac{\sqrt{9-p^2} + 1}{2})$ , the slope of $XP$ is $\frac{\sqrt{9-p^2} - 1}{p - 2}$ , and the perpendicular bisector of $XP$ is $y = -\frac{p - 2}{\sqrt{9-p^2} - 1}(x - \frac{p + 2}{2}) + \frac{\sqrt{9-p^2} + 1}{2}$ or $y = \frac{px - 2x - 2}{1 - \sqrt{9 - p^2}}$ .

The top part of the ellipse follows the minimum height of each $x$ -value over all $p$ -values, which occurs when $\frac{dy}{dp} = 0$ . Therefore, $\frac{dy}{dp} = \frac{9x - 2px - 2p - x\sqrt{9 - p^2}}{\sqrt{9 - p^2}(1 - \sqrt{9 - p^2})^2} = 0$ , which solves to $p = \frac{6x(3x + 3 - \sqrt{-x^2 + 2x + 1})}{5x^2 + 8x + 4}$ .

Substituting $p = \frac{6x(3x + 3 - \sqrt{-x^2 + 2x + 1})}{5x^2 + 8x + 4}$ into $y = \frac{px - 2x - 2}{1 - \sqrt{9 - p^2}}$ and simplifying gives $5x^2 + 8y^2 - 4xy - 8x - 4y - 4 = 0$ , so $A + B + C + D + E + F = 5 + 8 - 4 - 8 - 4 - 4 = \boxed{-7}$ .