If the two tangents drawn on hyperbola frac{{ | vedclass

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Question

(a) 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){\rm{ }}\

Vedclass · Accepted Answer

${y^2} + {b^2} = {c^2}({x^2} - {a^2})$

Vedclass · Answer

(a) Let $(h,k)$ be the point of intersection.

By $S{S_1} = {T^2}$, $\left( {\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} - 1} \right){\rm{ }}\left( {\frac{{{h^2}}}{{{a^2}}} - \frac{{{k^2}}}{{{b^2}}} - 1} \right) = {\left[ {\frac{{hx}}{{{a^2}}} - \frac{{ky}}{{{b^2}}} - 1} \right]^2}$

${x^2}\left[ {\frac{{{h^2}}}{{{a^4}}} - \frac{{{k^2}}}{{{a^2}{b^2}}} - \frac{1}{{{a^2}}} - \frac{{{h^2}}}{{{a^4}}}} \right] - {y^2}\left[ {\frac{{{h^2}}}{{{a^2}{b^2}}} - \frac{{{k^2}}}{{{b^4}}} - \frac{1}{{{b^2}}} + \frac{{{k^2}}}{{{b^4}}}} \right] + ... = 0$

We know that, ${m_1}{m_2} = \frac{{{\rm{Coefficent}}\,\,{\rm{of}}\,\,{x^2}}}{{{\rm{Coefficent}}\,\,{\rm{of}}\,\,{y^2}}}$

$ \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 $({y^2} + {b^2}) = {c^2}({x^2} - {a^2})$.

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

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