If a parabola with latus rectum $4a$ slides such that it touches the positive coordinate axes then find the locus of its focus.
If $(x_1,y_1)$ is a point in the first quadrant then the equation of parabola can be written as $(y-y_1)^2=4a(x-x_1)$ with focus, say, $(h,k)$.
If $(y-y_1)=Y, (x-x_1)=X$ then the parabola can be written as $Y^2=4aX$ with focus $(a,0)$ and a general point as $(at^2,2at)$.
So, an equation of tangent on this parabola is $tY=X+at^2$
In the solution, they have written that $h$ is the perpendicular distance of $(a,0)$ from the tangent $t_1Y=X+at_1^2$ and $k$ is the perpendicular distance of $(a,0)$ from the tangent $t_2Y=X+at_2^2$ with $t_1t_2=-1$ because the original parabola is sliding between the axes and they are perpendicular. Using this, I indeed get the final answer, which I am posting below for reference.
But I don't understand why $h$ and $k$ are perpendicular distances from $(a,0)$ to the tangents. I guess a diagram might help. I am trying to make different diagrams. One diagram is where the perpendiculars are dropped from $(h,k)$ onto the axes but they don't meet where the parabola is touching the axes. Is this scenario possible? If no, is there a way to visualize it convincingly or prove it algebraically?
Best Answer
Let the focus be $S(h,k)$. The feet of the perpendicular from focus on to the tangents are $(h,0)$ and $(0,k)$ both of which lie on tangent at vertex from the properties of parabola.
Hence equation of tangent at vertex is $\dfrac{x}{h}+\dfrac{y}{k}=1$
Distance of focus from tangent at vertex $=a$ yields $\dfrac{1}{x^2}+\dfrac{1}{y^2}= \dfrac{1}{a^2}$