(a) माना $(h,k)$ प्रतिच्छेद बिन्दु है।

$S{S_1} = {T^2}$ से,  $\left( {\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} – 1} \right){\rm{ }}\left( {\frac{{{h^2}}}{{{a^2}}} – \frac{{{k^2}}}{{{b^2}}} – 1} \right) = {\left[ {\frac{{hx}}{{{a^2}}} – \frac{{ky}}{{{b^2}}} – 1} \right]^2}$

${x^2}\left[ {\frac{{{h^2}}}{{{a^4}}} – \frac{{{k^2}}}{{{a^2}{b^2}}} – \frac{1}{{{a^2}}} – \frac{{{h^2}}}{{{a^4}}}} \right] – {y^2}\left[ {\frac{{{h^2}}}{{{a^2}{b^2}}} – \frac{{{k^2}}}{{{b^4}}} – \frac{1}{{{b^2}}} + \frac{{{k^2}}}{{{b^4}}}} \right] + … = 0$

हम जानते हैं, ${m_1}{m_2} = \frac{{{x^2}\,Coefficent}}{{{y^2}\ Coefficent}}$

$ \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}$ या $({y^2} + {b^2}) = {c^2}({x^2} – {a^2})$.