Projection of a Point on Conic and Distance of a Point from Conic Using Conic Intersection
Let's consider a Conic represented by General Quadratic Equations in 2 Variables as follows
\(Ax^2 + Bxy + Cy^2 + Dx + Ey + F=0\) ...(1)
The Projection of any given Point (\(x_0,y_0\)) on the Conic given by equation (1) above can be calculated using following steps
Find the Slope of the Conic. Since the Conic is an Implicit Function of \(x\) and \(y\) (i.e. \(F(x,y)=0\)), it's Partial Derivatives with respect to Variables \(x\) and \(y\) are calculated as
Find the Slope of Any Line Connecting the Point (\(x_0,y_0\)) to any Point (\(x,y\)) on the Conic is given below
Slope of Line \(={\Large \frac{y-y_0}{x-x_0}}\)
For Any Point of Projection (\(x,y\)) on the Conic of Point (\(x_0,y_0\)), the Slope of the Line Connecting the Point (\(x_0,y_0\)) to Point (\(x,y\)) is Perpendicular to Slope of the Conic at Point (\(x,y\)). Hence,
Slope of Line \(\times\) Slope of Conic \( = -{\Large \frac{2Ax + By + D}{2Cy + Bx + E}} \times {\Large \frac{y-y_0}{x-x_0}}=-1\)
Any Point of Projection (\(x,y\)) on the Conic of Point (\(x_0,y_0\)) Must Satisfy Both equations (1) and (5).
Hence, the Any Point of Projection (\(x,y\)) on the Conic of Point (\(x_0,y_0\)) can be calculated as Real Point(s) of Intersection Between 2 Conics given by equations (1) and (5).
Once all the Points of Projection (\(x,y\)) have been calculated, their Distances can be calculated from the Point (\(x_0,y_0\)). The one with the Minimum Value of Distance is the Distance Between the Point and the Conic.