1. ## locus problem

p311 q12
question : a variable line passing through the point (5,0) intersects the lines 3x-4y=0 and 3x+4y=0 at H and K respectively. Find the equation of the locus of the mid-point of HK.

2. Hello,

Originally Posted by afeasfaerw23231233
p311 q12
question : a variable line passing through the point (5,0) intersects the lines 3x-4y=0 and 3x+4y=0 at H and K respectively. Find the equation of the locus of the mid-point of HK.
Let $H(x_H, y_H)$ and $K(x_K, y_K)$

The equation of the variable line is : $y=ax+b$

But we know that it's passing through (5,0).
Thus $0=5a+b$
--> $b=-5a$

Therefore, the equation of the variable line is : $\boxed{y=ax-5a}$, a is the variable

So we know that :
\begin{aligned} y_H & = & ax_H-5a & (1) \\ y_K & = & ax_K-5a & (2) \end{aligned}
Because H and K are on this line.

Plus, H is on the line of equation $3x-4y=0$
And K is on the line of equation $3x+4y=0$
---> \begin{aligned} 3x_H-4y_H & = & 0 & (3) \\ 3x_K+4y_K & = & 0 & (4) \end{aligned}

By substituting (1) into (3), you will have $x_H$ with respect to a.
Replace this expression of $x_H$ in (1) to have $y_H$ with respect to a.

Do the same for $x_K$ and $y_k$.

Then, the midpoint of HK is $M \left(\frac{x_H+x_K}{2}, \frac{y_K+y_K}{2} \right)$, which will be coordinates with respect to a

Hope that helps... If you have questions about some steps, don't hesitate ^^

3. Originally Posted by afeasfaerw23231233
p311 q12
question : a variable line passing through the point (5,0) intersects the lines 3x-4y=0 and 3x+4y=0 at H and K respectively. Find the equation of the locus of the mid-point of HK.
$\frac{\left(x+\frac52\right)^2}{\left(\frac52\righ t)^2}- \frac{y^2}{\left(\frac{15}{8}\right)^2}=1$