Animated Solution for Mathematics - Circles: The sum of diameters of the circles that touch (i) the parabola $75x^2=64(5y-3)$ at the point $(\frac{8}{5},\frac{6}{5})$ and (ii) the y-axis, is equal to ______.

Welcome, future engineers! Today, we are going to dissect a problem that perfectly captures the elegance of coordinate geometry. We are exploring the delicate dance between a parabola and a circle.

Imagine you are standing on a coordinate plane, looking at a parabola defined by $75x^2=64(5y-3)$. It seems intimidating, but let's break it down.

First, we must see the parabola for what it truly is. The equation $75x^2=64(5y-3)$ is in a messy form. Let's bring it to the standard form $x^2=4a(y-k)$.

By dividing by $75$ and factoring out the $5$, we get:

This is an upward-opening parabola with its vertex at $(0,\frac{3}{5})$. Now, we are given a point $P=(\frac{8}{5},\frac{6}{5})$.

A quick check confirms this point lies on the parabola, as both sides of the equation yield $\frac{64}{25}$.

To find circles that touch the parabola at $P$, we need the tangent line at $P$. We use the powerful $T=0$ method.

For a parabola $x^2=4a(y-k)$, the tangent at $(x_1,y_1)$ is $x{x}_1=2a(y+y_1-2k)$. Substituting our values, we find the tangent line is:

Notice how beautifully this line passes through the origin! This is the "bridge" we need.

Here is where the magic happens. We need circles that touch this tangent line at $P$. Instead of guessing the center and radius, we use the "Family of Circles" equation:

Here, $(x_1,y_1)$ is our point $P$, and $L$ is our tangent line $3x-4y=0$. This equation represents all circles touching the line $L$ at $P$.

Expanding this, we get:

Now, the final constraint: the circle must also touch the y-axis. The y-axis is the line $x=0$.

If we substitute $x=0$ into our circle equation, we get a quadratic in $y$:

For the circle to touch the y-axis, this quadratic must have exactly one root. This means the discriminant $D=B^2-4AC$ must be zero.

Solving the following equation:

This yields two values for $\lambda$: $\lambda =\frac{2}{5}$ and $\lambda =-\frac{8}{5}$. These two values correspond to the two circles that satisfy our conditions.

For a circle touching the y-axis, the radius $r$ is simply the absolute value of the x-coordinate of the center. The x-center is $-\frac{1}{2}$ of the coefficient of $x$.

For $\lambda =\frac{2}{5}$, we find $r_1=1$. For $\lambda =-\frac{8}{5}$, we find $r_2=4$.

The diameters are $2r_1=2$ and $2r_2=8$. The sum of the diameters is:

And there you have it! A complex problem reduced to a beautiful, integer answer. Keep practicing, keep visualizing, and most importantly, keep falling in love with the logic!