Find the locus of intersection of tangents to an ellipse if the lines joining the points of contact to the centre be perpendicular.
Let the equation to the tangent be $$y=mx+\sqrt{a^2m^2+b^2} $$ This has to roots for $m$ that i.e $m_1$ and $m_2$.
perpendicular to this line passing through $(0,0)$ is $$my+x=0 $$ slope is $\frac{-1}{m}$
so for perpendiculars $$\frac{(-1)(-1)}{m_1m_2} =-1 $$ so $m_1m_2=-1 $
from equation of tangent $y=mx+\sqrt{a^2m^2+b^2} $
$$m^2(x^2-a^2)-2mxy+y^2-b^2=0 $$and hence locus is $$\frac{y^2-b^2}{x^2-a^2}=-1 $$ But this is not the correct answer. The answer is $b^4x^2+a^4y^2=a^2b^2(a^2+b^2) $. What's the error ?
Best Answer
Let $(x_1,y_1)$ be the generic point of the locus.
It is well known that $$\frac {xx_1}{a^2}+\frac {yy_1}{b^2}=1$$ represents the line passing through the points of contact of the tangents from $(x_1,y_1)$.
Now consider the equation $$\frac {x^2}{a^2}+\frac {y^2}{b^2}-\left(\frac {xx_1}{a^2}+\frac {yy_1}{b^2}\right)^2=0$$ It is satisfied by the coordinates of the center and the points of contact.
Since it can be written $$x^2 \left(\frac {x_1^2}{a^4}-\frac 1{a^2}\right) + y^2 \left(\frac {y_1^2}{b^4}-\frac 1{b^2}\right) + \frac {2\,x\,y\,x_1y_1}{a^2b^2}=0$$
it is quadratic homogeneous so represents a pair of lines (degenerate conic) , clearly the lines joining the points of contact to the centre.
It is not difficult to prove that the lines are mutually perpendicular iff the sum of the coefficients of $x^2$ and $y^2$ is zero, that is $$\frac {x_1^2}{a^4}+\frac {y_1^2}{b^4}=\frac 1{a^2}+\frac 1{b^2}$$