Conic Sections Question 109
Question: 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
Options:
A) $ m_1+m_2=\frac{24}{11} $
B) $ m_1m_2=\frac{20}{11} $
C) $ m_1+m_2=\frac{48}{11} $
D) $ m_1m_2=\frac{11}{20} $
Show Answer
Answer:
Correct Answer: B
Solution:
The line through (6,2) is $ y-2=m(x-6) $
therefore $ y=mx+2-6m $ Now from condition of tangency, $ {{(2-6m)}^{2}}=25m^{2}-16 $
therefore $ 36m^{2}+4-24m-25m^{2}+16=0 $
therefore $ 11m^{2}-24m+20=0 $ Obviously its roots are $ m_1 $ and $ m_2 $ , therefore $ m_1+m_2=\frac{24}{11} $ and $ m_1m_2=\frac{20}{11} $ .
BETA
