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If \mathrm{m_1} and \mathrm{ m_2} are the slopes of the tangents to the hyperbola \mathrm{ \frac{x^2}{25}-\frac{y^2}{16}=1}which pass through the point (6,2) if \mathrm{ m_1+m_2=\frac{24}{n}} then n is _____.

Option: 1

11


Option: 2

23


Option: 3

12


Option: 4

13


Answers (1)

best_answer

The line through (6,2) is

y-2=m(x-6) \Rightarrow y=m x+2-6 m

Now from condition of tangency, (2-6 m)^2=25 m^2-16

\begin{aligned} & \Rightarrow 36 m^2+4-24 m-25 m^2+16=0 \\ & \Rightarrow 11 m^2-24 m+20=0 \end{aligned}

Obviously its roots are m_1 and m_2, therefore

m_1+m_2=\frac{24}{11} \text { and } m_1 m_2=\frac{20}{11} \text {. }

Posted by

Irshad Anwar

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