Question

Prove that the straight $\frac{x}{a}-\frac{by}{b}=m$ and $\frac{x}{a}+\frac{y}{b}=\frac{1}{m}$ always meet on the hyperbola.

Answer 1

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Answer 2

Let the two given straight lines be

(1) $\frac{x}{a} - \frac{by}{b} = m$

(2) $\frac{x}{a} + \frac{y}{b} = \frac{1}{m}$

Assuming $b \neq 0$, the term $\frac{by}{b}$ simplifies to $y$. So, the first equation becomes:

(1) $\frac{x}{a} - y = m$

Now we have the system of equations:

(1) $\frac{x}{a} - y = m$

(2) $\frac{x}{a} + \frac{y}{b} = \frac{1}{m}$

Let the intersection point be $(x, y)$. We solve this system for $x$ and $y$.

Subtract equation (1) from equation (2):

$(\frac{x}{a} + \frac{y}{b}) - (\frac{x}{a} - y) = \frac{1}{m} - m$

$\frac{x}{a} + \frac{y}{b} - \frac{x}{a} + y = \frac{1}{m} - m$

$\frac{y}{b} + y = \frac{1-m^2}{m}$

$y(1 + \frac{1}{b}) = \frac{1-m^2}{m}$

$y(\frac{b+1}{b}) = \frac{1-m^2}{m}$

$y = \frac{b(1-m^2)}{m(b+1)}$

Now substitute the expression for $y$ back into equation (1) to find $x$:

$\frac{x}{a} = m + y$

$\frac{x}{a} = m + \frac{b(1-m^2)}{m(b+1)}$

$\frac{x}{a} = \frac{m^2(b+1) + b(1-m^2)}{m(b+1)}$

$\frac{x}{a} = \frac{m^2b + m^2 + b - bm^2}{m(b+1)}$

$\frac{x}{a} = \frac{m^2 + b}{m(b+1)}$

$x = \frac{a(m^2 + b)}{m(b+1)}$

The coordinates of the intersection point are $(x, y) = \left( \frac{a(m^2 + b)}{m(b+1)}, \frac{b(1-m^2)}{m(b+1)} \right)$.

We need to show that this point lies on a hyperbola. The phrase "always meet on the hyperbola" implies that the locus of the intersection point $(x, y)$ as $m$ varies is a hyperbola. We need to find the equation of this locus by eliminating $m$.

From the expressions for $x$ and $y$, we can write:

$\frac{x}{a} = \frac{m^2 + b}{m(b+1)}$

$\frac{y}{b} = \frac{1-m^2}{m(b+1)}$

Let's try to manipulate these expressions to eliminate $m$.

Consider the sum and difference of $\frac{x}{a}$ and $\frac{y}{b}$:

$\frac{x}{a} + \frac{y}{b} = \frac{m^2 + b}{m(b+1)} + \frac{1-m^2}{m(b+1)} = \frac{m^2 + b + 1 - m^2}{m(b+1)} = \frac{b+1}{m(b+1)} = \frac{1}{m}$

This matches the second given equation, which is expected.

$\frac{x}{a} - \frac{y}{b} = \frac{m^2 + b}{m(b+1)} - \frac{1-m^2}{m(b+1)} = \frac{m^2 + b - (1-m^2)}{m(b+1)} = \frac{m^2 + b - 1 + m^2}{m(b+1)} = \frac{2m^2 + b - 1}{m(b+1)}$

This does not seem to directly lead to a simple hyperbola equation.

Let's reconsider the probable intended form of the question based on standard problems of this type. If the equations were $\frac{x}{a} - \frac{y}{b} = m$ and $\frac{x}{a} + \frac{y}{b} = \frac{1}{m}$, the solution would be much cleaner and lead to a standard hyperbola form. Let's assume there was a typo in the first equation and it should have been $\frac{x}{a} - \frac{y}{b} = m$.

Assuming the lines are:

(1') $\frac{x}{a} - \frac{y}{b} = m$

(2') $\frac{x}{a} + \frac{y}{b} = \frac{1}{m}$

Add (1') and (2'):

$(\frac{x}{a} - \frac{y}{b}) + (\frac{x}{a} + \frac{y}{b}) = m + \frac{1}{m}$

$\frac{2x}{a} = m + \frac{1}{m}$

$\frac{x}{a} = \frac{1}{2}(m + \frac{1}{m})$

Subtract (1') from (2'):

$(\frac{x}{a} + \frac{y}{b}) - (\frac{x}{a} - \frac{y}{b}) = \frac{1}{m} - m$

$\frac{2y}{b} = \frac{1}{m} - m$

$\frac{y}{b} = \frac{1}{2}(\frac{1}{m} - m)$

Now, square both expressions:

$(\frac{x}{a})^2 = \left(\frac{1}{2}(m + \frac{1}{m})\right)^2 = \frac{1}{4}(m^2 + 2 + \frac{1}{m^2})$

$(\frac{y}{b})^2 = \left(\frac{1}{2}(\frac{1}{m} - m)\right)^2 = \frac{1}{4}(\frac{1}{m^2} - 2 + m^2)$

Subtract the second squared expression from the first:

$(\frac{x}{a})^2 - (\frac{y}{b})^2 = \frac{1}{4}(m^2 + 2 + \frac{1}{m^2}) - \frac{1}{4}(m^2 - 2 + \frac{1}{m^2})$

$= \frac{1}{4} [(m^2 + 2 + \frac{1}{m^2}) - (m^2 - 2 + \frac{1}{m^2})]$

$= \frac{1}{4} [m^2 + 2 + \frac{1}{m^2} - m^2 + 2 - \frac{1}{m^2}]$

$= \frac{1}{4} [4]$

$= 1$

So, we get the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. This is the equation of a hyperbola. The intersection point $(x, y)$ satisfies this equation regardless of the value of $m$. Thus, the intersection point always lies on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

This result strongly suggests that the first equation in the question was intended to be $\frac{x}{a} - \frac{y}{b} = m$. If we strictly follow the given equation $\frac{x}{a} - \frac{by}{b} = m$, which simplifies to $\frac{x}{a} - y = m$, the resulting locus is not a standard hyperbola form $\frac{x^2}{A^2} \pm \frac{y^2}{B^2} = 1$ or $\frac{y^2}{B^2} - \frac{x^2}{A^2} = 1$. It is possible that the intended hyperbola is not centered at the origin or has a more complex form, but without further context or clarification, the most reasonable interpretation is that there was a typo in the first equation.

Assuming the intended question used the lines $\frac{x}{a} - \frac{y}{b} = m$ and $\frac{x}{a} + \frac{y}{b} = \frac{1}{m}$, the proof is as follows:

Let the equations of the lines be

L1: $\frac{x}{a} - \frac{y}{b} = m$

L2: $\frac{x}{a} + \frac{y}{b} = \frac{1}{m}$

To find the intersection point $(x, y)$, we solve the system of equations.

Adding L1 and L2 gives:

$(\frac{x}{a} - \frac{y}{b}) + (\frac{x}{a} + \frac{y}{b}) = m + \frac{1}{m}$

$\frac{2x}{a} = m + \frac{1}{m}$

$\frac{x}{a} = \frac{1}{2}(m + \frac{1}{m})$

Subtracting L1 from L2 gives:

$(\frac{x}{a} + \frac{y}{b}) - (\frac{x}{a} - \frac{y}{b}) = \frac{1}{m} - m$

$\frac{2y}{b} = \frac{1}{m} - m$

$\frac{y}{b} = \frac{1}{2}(\frac{1}{m} - m)$

To find the locus of the intersection point, we eliminate the parameter $m$. We square the expressions for $\frac{x}{a}$ and $\frac{y}{b}$:

$(\frac{x}{a})^2 = \frac{1}{4}(m + \frac{1}{m})^2 = \frac{1}{4}(m^2 + 2 + \frac{1}{m^2})$

$(\frac{y}{b})^2 = \frac{1}{4}(\frac{1}{m} - m)^2 = \frac{1}{4}(\frac{1}{m^2} - 2 + m^2)$

Subtracting the second squared equation from the first:

$(\frac{x}{a})^2 - (\frac{y}{b})^2 = \frac{1}{4}(m^2 + 2 + \frac{1}{m^2}) - \frac{1}{4}(m^2 - 2 + \frac{1}{m^2})$

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = \frac{1}{4} [(m^2 + 2 + \frac{1}{m^2}) - (m^2 - 2 + \frac{1}{m^2})]$

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = \frac{1}{4} [4]$

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

This equation is the standard equation of a hyperbola. Since the coordinates $(x, y)$ of the intersection point satisfy this equation for any value of $m$ (where $m \neq 0$), the intersection point always lies on this hyperbola.