Normals are drawn to the ellipse $x^2 + 2y^2 = 2$ from the point (2,3) . Find the equation of the curve on which the co-normal points lie.

Question

I know that the sum of eccentric angles of all the co-normal points is an odd multiple of $\pi$. But I just can't figure out how that'll work in this question...

Answer 1

Hint

Taking the derivative w.r.t $x$ we get:

$$2x+4yy'=0 \rightarrow y'=-\frac{x}{2y}$$

So the slope of the normal lines are

$$y'_n=\frac{2y_0}{x_0}, \quad \text{where} \quad x_0^2+2y_0^2=2$$

Those lines go through the point $(2,3)$ so the general equation is:

$$y=\left(\frac{2y_0}{x_0}\right)x+\frac{3x_0-4y_0}{x_0}$$

The conormal points are the intersections of the above line with the ellipse at $(x_0y_0)$:

$$ y_0=\left(\frac{2y_0}{x_0}\right)x_0+\frac{3x_0-4y_0}{x_0}\rightarrow y_0=\frac{3x_0}{4-x_0}$$

Now the conormal points fit:

\begin{cases} x_0^2+2y_0^2=2\\ y_0=\frac{3x_0}{4-x_0} \end{cases}