If {m₁} and {m₂} are slopes of tangents to hyperbola frac{{{x^2}}}{{25}} - frac{{{y^2}}}{{16}} =

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10-2. Parabola, Ellipse, Hyperbola

hard

If ${m_1}$ and ${m_2}$are the slopes of the tangents to the hyperbola $\frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{{16}} = 1$ which pass through the point $(6, 2)$, then

A

${m_1} + {m_2} = \frac{{24}}{{11}}$

B

${m_1}{m_2} = \frac{{20}}{{11}}$

C

${m_1} + {m_2} = \frac{{48}}{{11}}$

D

both $(a)$ and $(b)$

Solution

(d) The line through $(6,2)$ is

$y – 2 = m(x – 6)$

==> $y = mx + 2 – 6m$

Now from condition of tangency, ${(2 – 6m)^2} = 25{m^2} – 16$

==> $36{m^2} + 4 – 24m – 25{m^2} + 16 = 0$

==> $11{m^2} – 24m + 20 = 0$

Obviously its roots are ${m_1}$ and ${m_2}$, therefore

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

Standard 11

Mathematics

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