Question

Property:Locus of the foot of perpendicular from focus of an ellipse to any tangent is

Answer 1

Auxiliary circle

Answer 2

Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a > b$. The foci are at $F_1(ae, 0)$ and $F_2(-ae, 0)$, where $e$ is the eccentricity. The equation of a tangent to the ellipse with slope $m$ is $y = mx \pm \sqrt{a^2 m^2 + b^2}$.

Consider the focus $F_1(ae, 0)$. Let $P(x, y)$ be the foot of the perpendicular from $F_1$ to the tangent $y = mx + \sqrt{a^2 m^2 + b^2}$. The line segment $F_1 P$ is perpendicular to the tangent. The slope of the tangent is $m$. The slope of $F_1 P$ is $\frac{y - 0}{x - ae} = \frac{y}{x - ae}$. Since the lines are perpendicular, the product of their slopes is $-1$: $m \times \frac{y}{x - ae} = -1$ $my = -(x - ae)$ $x + my = ae$ (Equation 1)

The point $P(x, y)$ lies on the tangent $y = mx + \sqrt{a^2 m^2 + b^2}$. $mx - y = -\sqrt{a^2 m^2 + b^2}$ (Equation 2)

We want to find the locus of $P(x, y)$, so we need to eliminate $m$ from Equations 1 and 2. Square both equations: From Equation 1: $(x + my)^2 = (ae)^2 \implies x^2 + 2mxy + m^2 y^2 = a^2 e^2$ From Equation 2: $(mx - y)^2 = (-\sqrt{a^2 m^2 + b^2})^2 \implies m^2 x^2 - 2mxy + y^2 = a^2 m^2 + b^2$

Add the two squared equations: $(x^2 + 2mxy + m^2 y^2) + (m^2 x^2 - 2mxy + y^2) = a^2 e^2 + a^2 m^2 + b^2$ $x^2 + m^2 y^2 + m^2 x^2 + y^2 = a^2 e^2 + a^2 m^2 + b^2$ $x^2(1 + m^2) + y^2(1 + m^2) = a^2 e^2 + a^2 m^2 + b^2$ $(x^2 + y^2)(1 + m^2) = a^2 e^2 + a^2 m^2 + b^2$

We know that for an ellipse, $b^2 = a^2 (1 - e^2)$, so $a^2 e^2 = a^2 - b^2$. Substitute this into the equation: $(x^2 + y^2)(1 + m^2) = (a^2 - b^2) + a^2 m^2 + b^2$ $(x^2 + y^2)(1 + m^2) = a^2 + a^2 m^2$ $(x^2 + y^2)(1 + m^2) = a^2 (1 + m^2)$

Assuming $1 + m^2 \neq 0$ (which is always true for real $m$), we can divide both sides by $1 + m^2$: $x^2 + y^2 = a^2$

This is the equation of a circle centered at the origin $(0, 0)$ with radius $a$. This circle is known as the auxiliary circle of the ellipse.

If we started with the other focus $F_2(-ae, 0)$, the equations would be: $x - my = -ae$ (perpendicularity condition) $mx - y = -\sqrt{a^2 m^2 + b^2}$ (point on the tangent) Squaring and adding these equations gives: $(x - my)^2 + (mx - y)^2 = (-ae)^2 + (-\sqrt{a^2 m^2 + b^2})^2$ $x^2 - 2mxy + m^2 y^2 + m^2 x^2 - 2mxy + y^2 = a^2 e^2 + a^2 m^2 + b^2$ $x^2(1 + m^2) + y^2(1 + m^2) - 4mxy = a^2 e^2 + a^2 m^2 + b^2$ This does not seem right. Let's recheck the equation of the perpendicular line from $F_2(-ae, 0)$. The slope of the tangent is $m$. The slope of the perpendicular line from $F_2(-ae, 0)$ is $-1/m$. The equation of the perpendicular line is $y - 0 = -\frac{1}{m} (x - (-ae))$, so $y = -\frac{1}{m} (x + ae)$, which is $my = -(x + ae)$, or $x + my = -ae$.

So the system of equations for the foot of the perpendicular from $F_2$ is: 1') $x + my = -ae$ 2') $mx - y = -\sqrt{a^2 m^2 + b^2}$ (same tangent equation)

Square both equations: $(x + my)^2 = (-ae)^2 \implies x^2 + 2mxy + m^2 y^2 = a^2 e^2$ $(mx - y)^2 = (-\sqrt{a^2 m^2 + b^2})^2 \implies m^2 x^2 - 2mxy + y^2 = a^2 m^2 + b^2$

Add the two squared equations: $(x^2 + 2mxy + m^2 y^2) + (m^2 x^2 - 2mxy + y^2) = a^2 e^2 + a^2 m^2 + b^2$ $x^2(1 + m^2) + y^2(1 + m^2) = a^2 e^2 + a^2 m^2 + b^2$ $(x^2 + y^2)(1 + m^2) = a^2 e^2 + a^2 m^2 + b^2$ Using $a^2 e^2 = a^2 - b^2$: $(x^2 + y^2)(1 + m^2) = (a^2 - b^2) + a^2 m^2 + b^2$ $(x^2 + y^2)(1 + m^2) = a^2 + a^2 m^2$ $(x^2 + y^2)(1 + m^2) = a^2 (1 + m^2)$ $x^2 + y^2 = a^2$

In both cases (perpendicular from $F_1$ or $F_2$), the locus of the foot of the perpendicular is the auxiliary circle $x^2 + y^2 = a^2$.

The auxiliary circle is a circle with center at the center of the ellipse and radius equal to the semi-major axis $a$.