Question

The coordinates of a point are $(a \tan (\theta + \alpha), b \tan (\theta + \beta))$, where $\theta$ is variable; prove that the locus of a point which divides it in any given ratio is a hyperbola.

Answer 1

The locus of the point is a hyperbola

Answer 2

Let the coordinates of the point be $P = (a \tan (\theta + \alpha), b \tan (\theta + \beta))$.

Let the point which divides the segment joining the origin $O(0,0)$ and $P$ in the ratio $m:n$ be $Q(x,y)$. Using the section formula, the coordinates of $Q$ are:

$x = \frac{n(0) + m(a \tan (\theta + \alpha))}{m+n} = \frac{ma}{m+n} \tan (\theta + \alpha)$

$y = \frac{n(0) + m(b \tan (\theta + \beta))}{m+n} = \frac{mb}{m+n} \tan (\theta + \beta)$

Let $k = \frac{m}{m+n}$. Since the ratio is given, $k$ is a constant.

So, $x = ka \tan (\theta + \alpha)$ and $y = kb \tan (\theta + \beta)$.

From these equations, we have:

$\tan (\theta + \alpha) = \frac{x}{ka}$

$\tan (\theta + \beta) = \frac{y}{kb}$

We need to eliminate the variable $\theta$. Consider the difference of the angles $(\theta + \alpha)$ and $(\theta + \beta)$:

$(\theta + \alpha) - (\theta + \beta) = \alpha - \beta$.

Let $A = \theta + \alpha$ and $B = \theta + \beta$. Then $A - B = \alpha - \beta$.

We use the tangent subtraction formula:

$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

Substituting the expressions for $\tan A$ and $\tan B$:

$\tan(\alpha - \beta) = \frac{\frac{x}{ka} - \frac{y}{kb}}{1 + \frac{x}{ka} \cdot \frac{y}{kb}}$

$\tan(\alpha - \beta) = \frac{\frac{xb - ya}{kab}}{\frac{kab + xy}{kab}}$

$\tan(\alpha - \beta) = \frac{xb - ya}{kab + xy}$

Let $C = \tan(\alpha - \beta)$. Since $\alpha$ and $\beta$ are constants, $C$ is a constant.

$C(kab + xy) = xb - ya$

$Ckab + Cxy = xb - ya$

Rearranging the terms to get the equation of the locus:

$Cxy - xb + ay + Ckab = 0$

This is a general second-degree equation in $x$ and $y$ of the form $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, where $A=0$, $B=C$, $C=0$, $D=-b$, $E=a$, and $F=Ckab$.

The type of conic section represented by the general second-degree equation is determined by the discriminant $B^2 - 4AC$. In this case, the discriminant is $C^2 - 4(0)(0) = C^2$.

If $C = \tan(\alpha - \beta) \neq 0$, i.e., $\alpha - \beta$ is not an integer multiple of $\pi$, then $C^2 > 0$. A conic section with a positive discriminant is a hyperbola or a pair of intersecting lines.

The equation of the locus is $Cxy - bx + ay + Ckab = 0$. If $C \neq 0$, we can rewrite this as $xy - \frac{b}{C}x + \frac{a}{C}y + kab = 0$. This equation can be factored as $(x + \frac{a}{C})(y - \frac{b}{C}) - \frac{ab}{C^2} + kab = 0$. $(x + \frac{a}{C})(y - \frac{b}{C}) = \frac{ab}{C^2} - kab = \frac{ab - kabC^2}{C^2} = \frac{ab(1 - kC^2)}{C^2}$. Let $X = x + \frac{a}{C}$ and $Y = y - \frac{b}{C}$. This is a translation of axes. The equation becomes $XY = \frac{ab(1 - kC^2)}{C^2}$.

If $a \neq 0$, $b \neq 0$, $C \neq 0$, and $1 - kC^2 \neq 0$, this is the equation of a rectangular hyperbola with asymptotes $X=0$ and $Y=0$.

If $C = \tan(\alpha - \beta) = 0$, then $\alpha - \beta = n\pi$ for some integer $n$. In this case, $\tan(\theta + \alpha) = \tan(\theta + \beta)$. The original equations become $x = ka \tan(\theta + \alpha)$ and $y = kb \tan(\theta + \alpha)$. So, $\frac{x}{ka} = \frac{y}{kb}$, which implies $kbx = kay$, or $bx - ay = 0$. This is the equation of a straight line passing through the origin. A straight line is considered a degenerate hyperbola (a pair of intersecting lines where the intersection point is at infinity).

If $a=0$, the original point is $(0, b \tan(\theta + \beta))$. The locus of the dividing point is $(0, kb \tan(\theta + \beta))$. If $\theta$ is variable, $\tan(\theta+\beta)$ can take any real value, so the locus is the entire y-axis $x=0$. This is a degenerate hyperbola. If $b=0$, the original point is $(a \tan(\theta + \alpha), 0)$. The locus of the dividing point is $(ka \tan(\theta + \alpha), 0)$. This is the entire x-axis $y=0$. This is a degenerate hyperbola.

In general, assuming $a \neq 0$, $b \neq 0$, and $\alpha - \beta$ is not an integer multiple of $\pi$, the locus is a non-degenerate hyperbola. The equation $Cxy - bx + ay + Ckab = 0$ represents a hyperbola.