• # question_answer If the two tangents drawn on hyperbola $\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1$ in such a way that the product of their gradients is ${{c}^{2}}$, then they intersects on the curve A)            ${{y}^{2}}+{{b}^{2}}={{c}^{2}}({{x}^{2}}-{{a}^{2}})$                    B)            ${{y}^{2}}+{{b}^{2}}={{c}^{2}}({{x}^{2}}+{{a}^{2}})$ C)            $a{{x}^{2}}+b{{y}^{2}}={{c}^{2}}$    D)            None of these

Let $(h,k)$be the point of intersection. By$S{{S}_{1}}={{T}^{2}}$,             $\left( \frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}-1 \right)\text{ }\left( \frac{{{h}^{2}}}{{{a}^{2}}}-\frac{{{k}^{2}}}{{{b}^{2}}}-1 \right)={{\left[ \frac{hx}{{{a}^{2}}}-\frac{ky}{{{b}^{2}}}-1 \right]}^{2}}$                   Þ${{x}^{2}}\left[ \frac{{{h}^{2}}}{{{a}^{4}}}-\frac{{{k}^{2}}}{{{a}^{2}}{{b}^{2}}}-\frac{1}{{{a}^{2}}}-\frac{{{h}^{2}}}{{{a}^{4}}} \right]-{{y}^{2}}\left[ \frac{{{h}^{2}}}{{{a}^{2}}{{b}^{2}}}-\frac{{{k}^{2}}}{{{b}^{4}}}-\frac{1}{{{b}^{2}}}+\frac{{{k}^{2}}}{{{b}^{4}}} \right]+...=0$                   We know that, ${{m}_{1}}{{m}_{2}}=\frac{\text{Coefficent}\,\,\text{of}\,\,{{x}^{2}}}{\text{Coefficent}\,\,\text{of}\,\,{{y}^{2}}}$                   $\Rightarrow$${{m}_{1}}{{m}_{2}}=\frac{\frac{{{k}^{2}}}{{{a}^{2}}{{b}^{2}}}+\frac{1}{{{a}^{2}}}}{\frac{{{h}^{2}}}{{{a}^{2}}{{b}^{2}}}-\frac{1}{{{b}^{2}}}}={{c}^{2}}$                    $\Rightarrow$$\left( \frac{{{k}^{2}}+{{b}^{2}}}{{{h}^{2}}-{{a}^{2}}} \right)={{c}^{2}}$ or $({{y}^{2}}+{{b}^{2}})={{c}^{2}}({{x}^{2}}-{{a}^{2}})$.