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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.

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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 $y$.

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  • $\begingroup$ understood, just one query why did you equate $ T^2=4S^'S $ and not $ T^2=S^'S $ $\endgroup$
    – goku
    Oct 4 '20 at 17:08
  • $\begingroup$ @goku the factor of 4 was a mistake. I have corrected it now. Thanks. $\endgroup$
    – Z Ahmed
    Oct 5 '20 at 4:13

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