A pair of tangents to conic $ax^2+by^2=1$ intercepts a constant distance 2k on the y-axis. Prove that locus of their intersection is the conic.
$ax^2(ax^2+by^2-1)=bk^2(ax^2-1)^2$
I tried by introducing two tangents with slopes $m_1$ and $m_2$ and finding their $y$ intercept and equating it to 2k but not sure what to do after it.
Any help appreciated.
Best Answer
The combines equation of pair of tangents to the conic is given by $T^2=S'S$: $$(axx'+byy'-1)^2= S'(ax^2+by^2-1), S'=(ax'^2+by'^2-1)~~~(1)$$ Let them cut $y$axis, put $x=0$ to get a quadratic in $y$ as $$(byy'-1)^2=S'(by^2-1)=0 \implies (b^2y'^2-S'b) y^2-2by'y-S'=0$$ This gives $$y_1+y_2=2by'/(b^2y'^2-bS'), y_1y_2=-S'/(b^2y'^2-bS')$$ From these eqns we can get: $$2k=y_1-y_2=\sqrt{(y_1+y_2)^2-y_1y_2} ~~~(2)$$ Get this equation (2) and put $x'=x$ and $y'=y$, to get the required locus of $(x', y')$.
Please note the the locus will not be conic as it would not be a quadratic of $x$ and $y$.