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If the two tangents drawn on hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1} in such a way that the product of their gradients is \mathrm{c^2}, then they intersects on the curve

Option: 1

\mathrm{y^2+b^2=c^2\left(x^2-a^2\right)}


Option: 2

\mathrm{y^2+b^2=c^2\left(x^2+a^2\right)}


Option: 3

\mathrm{a x^2+b y^2=c^2}


Option: 4

\mathrm{\text { None of these }}


Answers (1)

best_answer

Let (h, k) be the point of intersection. By \mathrm{S S_1=T^2,}

\mathrm{ \begin{aligned} & \left(\frac{x^2}{a^2}-\frac{y^2}{b^2}-1\right)\left(\frac{h^2}{a^2}-\frac{k^2}{b^2}-1\right)=\left[\frac{h x}{a^2}-\frac{k y}{b^2}-1\right]^2 \\\\ & \Rightarrow\left[\frac{h^2}{a^4}-\frac{k^2}{a^2 b^2}-\frac{1}{a^2}-\frac{h^2}{a^4}\right] x^2-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]+\ldots=0} \end{aligned} }

\mathrm{\text { We know that, } m_1 m_2=\frac{\text { Coefficient of } x^2}{\text { Coefficient of } y^2}}

\mathrm{\begin{gathered} \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 \text { or }\left(y^2+b^2\right)=c^2\left(x^2-a^2\right) . \end{gathered}}

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Rishi

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