Animated Solution for Mathematics - Conic Sections: Match the following : $(3,0)$ is the point from which three normals are drawn to the parabola $y^2=4x$ which meet the parabola in the points $P,Q$ and $R$. Then

Imagine standing at the point $(3,0)$ on the Cartesian plane, staring at the elegant curve of the parabola $y^2=4x$. This curve is the locus of points equidistant from a focus and a directrix.

We are interested in the normals to this parabola. A normal is a line perpendicular to the tangent at a specific point. Drawing three normals from a single point $(3,0)$ to this parabola allows us to find three distinct paths that hit the curve at right angles.

To begin, we utilize the general equation of a normal to the parabola $y^2=4ax$ with slope $m$:

Comparing $y^2=4x$ with $y^2=4ax$, we identify that $a=1$. This simplifies our equation to:

Since these normals must pass through the point $(3,0)$, we substitute $x=3$ and $y=0$ into the equation:

Simplifying this expression, we obtain:

Factoring the equation $m(1-m^2)=0$, we find three distinct slopes: $m=0$, $m=1$, and $m=-1$. These represent the slopes of the three normals.

The parametric coordinates of a point on the parabola $y^2=4ax$ where the normal has slope $m$ are given by $(a{m}^2,-2am)$. With $a=1$, these points are $(m^2,-2m)$.

We calculate the vertices $P,Q,$ and $R$ as follows:

For $m=0$, we get $P(0,0)$. For $m=1$, we get $R(1,-2)$. * For $m=-1$, we get $Q(1,2)$.

The vertices of our triangle are $(0,0)$, $(1,2)$, and $(1,-2)$. Note that $Q$ and $R$ share the same x-coordinate, meaning the side $QR$ is a vertical line segment.

The area of $\Delta PQR$ is calculated using the base $QR$ and the height from $P$. The length of the base $QR$ is $|2-(-2)|=4$.

The height of the triangle, which is the perpendicular distance from $P(0,0)$ to the line $x=1$, is $1$. Thus, the area is:

The centroid $G$ is the average of the coordinates:

Because the triangle is symmetric about the x-axis, the circumcentre $C$ must lie on the x-axis at some point $(x,0)$. The distance from $C$ to $P(0,0)$ must equal the distance from $C$ to $Q(1,2)$.

Setting $C{P}^2=C{Q}^2$, we get:

Expanding this, we have $x^2=x^2-2x+1+4$, which simplifies to $2x=5$, or $x=\frac{5}{2}$. Thus, the circumcentre is $(\frac{5}{2},0)$.

The radius of the circumcircle is the distance from $C(\frac{5}{2},0)$ to $P(0,0)$, which is $\frac{5}{2}$.