Animated Solution for Mathematics - Conic Sections: If $y=m_{1}x+c_1$ and $y=m_{2}x+c_2,m_1\ne m_2$ are two common tangents of circle $x^2+y^2=2$ and parabola $y^2=x$, then the value of $8|m_1{m}_2|$ is equal to

Welcome, fellow traveler of the mathematical landscape. Today, we stand before two elegant curves: the circle $x^2+y^2=2$ and the parabola $y^2=x$.

They are distinct, yet they share a common destiny—the common tangents. Imagine you are a surveyor trying to draw a line that kisses both these curves simultaneously. How do we find such a line? Let us embark on this journey.

Our first step is to understand the parabola $y^2=x$. This is a rightward-opening parabola.

In the standard form $y^2=4ax$, we see that $4a=1$, which means $a=\frac{1}{4}$. A line $y=mx+c$ is tangent to this parabola if and only if $c=\frac{a}{m}$.

Substituting our value of $a$, the equation of any tangent to this parabola is:

This is our bridge. It is a line that is guaranteed to touch the parabola. Now, we need to ensure this same line also touches our circle.

The circle $x^2+y^2=2$ is centered at the origin $(0,0)$ with a radius $r=\sqrt{2}$. For a line to be tangent to a circle, the perpendicular distance from the center of the circle to the line must be exactly equal to the radius.

Let us rewrite our tangent equation in the standard form:

The distance $d$ from the origin $(0,0)$ to this line is given by the formula $d=\frac{|A{x}_1+B{y}_1+C|}{\sqrt{A^2+B^2}}$. Plugging in our values, we get:

Setting this equal to the radius $\sqrt{2}$, we have:

Now, the tension builds. We have an equation, but it is wrapped in a modulus and a square root. Let us simplify.

Squaring both sides, we get:

Cross-multiplying, we find $1=32m^2(m^2+1)$, which expands to:

This is a bi-quadratic equation, but do not fear! Let $t=m^2$. Our equation becomes $32t^2+32t-1=0$.

Using the quadratic formula:

The discriminant is $\sqrt{1024+128}=\sqrt{1152}=24\sqrt{2}$. Thus:

Since $m$ is a real slope, $m^2$ must be positive. The value $\frac{-4-3\sqrt{2}}{8}$ is clearly negative, so we reject it. We are left with:

We are asked to find $8|m_1{m}_2|$. Since $m_1$ and $m_2$ are the slopes of the two common tangents, they are the two roots of our quadratic in $m$.

The two slopes are $m=\pm \sqrt{\frac{3\sqrt{2}-4}{8}}$. Their product is $m_1{m}_2=-m^2=-\frac{3\sqrt{2}-4}{8}$.

The magnitude $|m_1{m}_2|$ is just $m^2$. So:

The elegance of the cancellation is breathtaking. We have arrived at the solution: $3\sqrt{2}-4$. You have mastered the geometry and the algebra. Well done!