If a parabola with latus rectum $4a$ slides such that it touches the positive coordinate axes then find the locus of its focus. 
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?
 A: 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 $\dfrac{1}{x^2}+\dfrac{1}{y^2}= \dfrac{1}{a^2}$
A: 
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?

Not only possible, but necessary.
One way that a Cartesian coordinate system is often explained is by constructing a rectangle with sides parallel to the axes,
placing one vertex of the rectangle at the origin and placing the diagonally opposite vertex at the point whose coordinates you want to determine.
The lengths of the sides of the rectangle -- which are perpendiculars dropped from the point in question to the axes -- are the coordinates of the point if it is in the first quadrant.
In short, the coordinates of a point in the first quadrant of a Cartesian coordinate system are literally the lengths of perpendiculars dropped from that point to the axis. There's no other way for it to be.
Since your two tangent lines are the $x$ and $y$ axes of the $x,y$ Cartesian coordinate system, the $(x,y)$ coordinates of the focus of the parabola must be the lengths of the perpendiculars dropped to those two lines.
This is not because the point $(x,y)$ is the focus; it's simply because it's a point in the first quadrant of that coordinate system.
Since the vertex is the only point $P$ on a parabola where a line through $P$ and the focus is perpendicular to the tangent at $P$, and since the parabola cannot be tangent to one axis at its vertex while simultaneously being tangent to the other axis, you will never find that the perpendicular meets the axis at the tangent point.
A: $$h=\left|\frac{a+at_1^2}{\sqrt{1+t_1^2}}\right|$$Squaring and rearranging,$$t_1^2=\frac{h^2}{a^2}-1$$Similarly,$$t_2^2=\frac{k^2}{a^2}-1$$Now, $t_1^2t_2^2=1$, so,$$\left(\frac{h^2}{a^2}-1\right)\left(\frac{k^2}{a^2}-1\right)=1\\\implies\frac{h^2k^2}{a^4}-\frac{h^2}{a^2}-\frac{k^2}{a^2}=0$$Therefor, the required locus is$$\frac1{x^2}+\frac1{y^2}=\frac1{a^2}$$
