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  • Tangents drawn from the point (c, d) to the hyperbola <img alt="\mathrm{ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20%5Cfrac%7Bx%5E2

Tangents drawn from the point (c, d) to the hyperbola \mathrm{ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1} make angles \mathrm{ \alpha} and \mathrm{\beta} with the x-axis. If \mathrm{\tan \alpha \tan \beta=1} then find the value of \mathrm{ c^2-d^2.}

Option: 1

\mathrm{a^2-b^2}


Option: 2

\mathrm{(a b)^2}


Option: 3

\mathrm{a-b}


Option: 4

\mathrm {a^2+b^2}


Answers (1)

best_answer

One of the equations of tangents to the hyperbola having slope m is \mathrm{y=m x+\sqrt{a^2 m^2-b^2}}. It passes through (c, d). So,

\mathrm{ \begin{aligned} & d=m c+\sqrt{a^2 m^2-b^2} \\\\ & \text { or }(d-m c)^2=a^2 m^2-b^2 \\\\ & \Rightarrow\left(c^2-a^2\right) m^2-2 c d m+d^2+b^2=0 \end{aligned} }

or \mathrm{(d-m c)^2=a^2 m^2-b^2}

\mathrm{ \Rightarrow\left(c^2-a^2\right) m^2-2 c d m+d^2+b^2=0 }
or Product of roots \mathrm{ =m_1 m_2=\frac{d^2+b^2}{c^2-a^2}}

\mathrm{ \Rightarrow \tan \alpha \tan \beta=\frac{d^2+b^2}{c^2-a^2}=1 \Rightarrow d^2+b^2=c^2-a^2 }

or \mathrm{c^2-d^2=a^2+b^2}

Posted by

Irshad Anwar

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