Animated Solution for Mathematics - Conic Sections: Let P be a point on the parabola, $x^2=4y$. If the distance of P from the centre of the circle, $x^2+y^2+6x+8=0$ is minimum, then the equation of the tangent to the parabola at P, is :

Welcome, my dear student! Today, we are going to unravel a problem that beautifully bridges the gap between coordinate geometry and the intuitive nature of physics. We are looking for the point $P$ on the parabola $x^2=4y$ that is closest to the center of the circle $x^2+y^2+6x+8=0$.

This is not just an algebraic exercise; it is a journey into the heart of curves.

First, let us tame the circle. The given equation is $x^2+y^2+6x+8=0$.

To find the center, we complete the square. We group the $x$ terms: $(x^2+6x)+y^2=-8$. Adding $9$ to both sides gives $(x^2+6x+9)+y^2=-8+9$, which simplifies to:

This is a circle centered at $C(-3,0)$ with a radius of $1$. Visualize this center on your coordinate plane; it is sitting on the negative $x$-axis.

Now, here is the core of the problem. You might be tempted to use the distance formula and minimize it using calculus, but there is a more elegant way.

The shortest distance from a point to a curve always lies along the normal to the curve at that point. Imagine a circle centered at $C$ expanding until it touches the parabola.

The moment it touches, the radius of that circle is the normal to the parabola. Therefore, the normal at $P$ must pass through $C(-3,0)$. This is our golden key!

For the parabola $x^2=4y$, we know that $4a=4$, so $a=1$. We can represent any point $P$ on this parabola as $(2t,t^2)$.

This is a powerful tool because it reduces our search for a point $(x,y)$ to finding a single value $t$. Now, let us find the slope of the tangent at $P$.

Differentiating $x^2=4y$ with respect to $x$, we get $2x=4\frac{dy}{dx}$, which means:

At our point $P(2t,t^2)$, the slope of the tangent $m_T$ is $\frac{2t}{2}=t$. Since the normal is perpendicular to the tangent, its slope $m_N$ is $-\frac{1}{t}$.

We now write the equation of the normal at $P(2t,t^2)$ using the point-slope form:

We know this normal passes through $C(-3,0)$. Substituting these values:

Multiplying by $-t$, we get $t^3=-3-2t$, or $t^3+2t+3=0$. This looks intimidating, but as we discussed, look for the simple root. Testing $t=-1$:

It works!

With $t=-1$, our point $P$ is $(2(-1),(-1{)}^2)=(-2,1)$.

The slope of the tangent at this point is $m_T=t=-1$. Now, we simply write the equation of the tangent line passing through $(-2,1)$ with slope $-1$:

This simplifies to $y-1=-x-2$, or $x+y+1=0$.

We have arrived! The elegance of this solution lies in the fact that we never had to perform complex minimization; we simply used the geometric property of the normal. Keep this intuition with you, and you will conquer any geometry problem that comes your way!