Finding locus of point of intersection of pair of tangents

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.

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 $$y$$.
• understood, just one query why did you equate $T^2=4S^'S$ and not $T^2=S^'S$