Find the locus of points whose distances from the line $y=\sqrt3x$ and x-axis are equal.

Question

Find the locus of points whose distances from the line$\hspace{0.2cm}$ $y=\sqrt3x$$\hspace{0.2cm}$ and x-axis are equal.

My solution:I start with the following $$\frac{|\sqrt3x_1-y_1|}{2}=\frac{|y_1|}{1}$$ After squaring both sides and simplification I got finally $$3x_1^2-3y_1^2-2\sqrt3x_1y_1=0$$ By comparing with $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ In my problem I have $$B^2-4AC=12+36=48>0$$ So it should be a hyperbola,infact a rectangular hyperbola(because $A+C=0$).

My Problem:If we think geometrically then the line$\hspace{0.2cm}$ $y=\sqrt3x$ $\hspace{0.2cm}$has angle$\hspace{0.2cm}$ $60^o$ $\hspace{0.2cm}$with x-axis.So geometrically the locus off all the points whose distances from the line $\hspace{0.2cm}$$y=\sqrt3x$ $\hspace{0.2cm}$and x-axis are equal should be a line making angle $\hspace{0.2cm}$$30^o$$\hspace{0.2cm}$ with x-axis. So please someone help me to find out the locus.

Thanks.

Answer 1

No, the answer is two lines $y=\frac{x}{\sqrt 3},y=-\sqrt 3x$ because $$0=3x^2-3y^2-2\sqrt 3xy=(x-\sqrt 3y)(\sqrt 3y+3x).$$

Answer 2

$$ax^2+2hxy+by^2+2gx+2fy+c=0$$ will be degenerate conic if

$$\begin{vmatrix} a&h&g\\h&b&f\\g&f&c\end{vmatrix}=0$$

and it will represent a pair of straight lines real or imaginary according as $ab-h^2\le0$ or $>0$

Ellipse, Parabola & Hyperbola are all non-degenerate conics