Animated Solution for Mathematics - Conic Sections: Two common tangents to the circle $x^2+y^2=2a^2$ and parabola $y^2=8ax$ are

We are tasked with finding the common tangents to the circle $x^2+y^2=2a^2$ and the parabola $y^2=8ax$. The circle is centered at the origin $(0,0)$ with a radius $r=\sqrt{2}a$.

The parabola $y^2=8ax$ opens to the right. We seek lines that satisfy the tangency conditions for both geometric entities simultaneously.

Any line tangent to a parabola of the form $y^2=4Ax$ can be expressed as:

Comparing our parabola $y^2=8ax$ to the standard form, we identify $4A=8a$, which implies $A=2a$.

Substituting this into our general tangent equation, we obtain the family of lines:

For this line to be tangent to the circle $x^2+y^2=2a^2$, the perpendicular distance from the center $(0,0)$ to the line $mx-y+\frac{2a}{m}=0$ must equal the radius $r=\sqrt{2}a$.

Using the perpendicular distance formula $d=\frac{|A{x}_1+B{y}_1+C|}{\sqrt{A^2+B^2}}$, we set up the following equation:

Simplifying the expression, we get:

Assuming $a>0$, we cancel $a$ from both sides and square the resulting equation to eliminate the radical and the modulus:

This simplifies to the biquadratic equation:

By substituting $u=m^2$, we solve the quadratic $u^2+u-2=0$, which factors as $(u+2)(u-1)=0$. Since $m^2$ must be positive, we discard $u=-2$ and accept $m^2=1$, yielding $m=\pm 1$.

Substituting $m=1$ and $m=-1$ back into the tangent equation $y=mx+\frac{2a}{m}$, we find the two common tangents:

These can be expressed compactly as $y=\pm (x+2a)$. You have successfully determined the lines that bridge these two geometric worlds.