If the line $mx + ny = 3$ is normal to the hyperbola $x^2 – y^2 = 1$, then evaluate $\frac{1}{m^2}+\frac{1}{n^2}$.
I compared given equation of normal to equation of normal at parametric point i.e $x\sin\theta+y=2\tan\theta$ and obtained $\frac{1}{m^2}-\frac{1}{n^2}=\frac{4}{9}$ but can't use it to obtain $\frac{1}{m^2}+\frac{1}{n^2}$. Any suggestion?
There must be a flaw in the text, expression $\frac{1}{m^2}+\frac{1}{n^2}$, should have been $\frac{1}{m^2}-\frac{1}{n^2}$.
Let us see why:
First of all, I confirm the expression given by @Mathematician :
$$\frac{1}{m^2}-\frac{1}{n^2}=\frac{4}{9} \ \ \ (1)$$
Proof below.
In fact, as the question is formulated, the only reasonable answer should have been "$\frac{1}{m^2}+\frac{1}{n^2}$ is this constant". But due to (1),
$$\frac{1}{m^2}+\frac{1}{n^2}=\frac{4}{9}+\frac{2}{n^2}$$
which is a variable quantity.
Proof of (1): The equation of the tangent line $(T)$ at $(x_0,y_0)$ of the hyperbola $x^2-y^2=1$ is $xx_0-yy_0=1 \ \ (2)$.
Straight line $(D_{mn})$ with equation $mx+ny=1$ must fillful two conditions:
$(D_{mn})$ and $(T)$ should be orthogonal, i.e., their normal vectors $(x_0,-y_0)$ and $(m,n)$ are orthogonal, i.e., $mx_0-ny_0=0 \ \ (3)$
$(D_{mn})$ must pass through point $(x_0,y_0)$, i.e., $mx_0+ny_0=3 \ \ (4)$
Solving the system constituted by (3) and (4) for $x_0$ and $y_0$, we get:
$$x_0=\frac{3}{2m} \ \ and \ \ y_0=\frac{3}{2n} $$
Finally using constraint $x_0^2-y_0^2=1$, we obtain condition (1).