Tangents are drawn to the hyperbola \frac{x^{2}}{9} \textrm{ }\textrm{ } - \textrm{ }\textrm{ } \frac{y^{2}}{4} \textrm{ }\textrm{ } = \textrm{ }\textrm{ } 1 , parallel to the straight line 2x – y = 1. The points of contact of the tangent on the hyperbola are

Engineering

Mathematics

Tangent to Hyperbola

Question

Tangents are drawn to the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ , parallel to the straight line 2x – y = 1. The points of contact of the tangent on the hyperbola are

$(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}})$

$(3 \sqrt{3}, - 2 \sqrt{2})$

$(\frac{- 9}{2 \sqrt{2}}, \frac{- 1}{\sqrt{2}})$

$(- 3 \sqrt{3}, 2 \sqrt{2})$

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If line y = mx + c is tangent at (x₁, y₁) then contact points is $(\frac{- a^{2} m}{c}, \frac{- b^{2}}{c})$

Equation of tangent of slope m is y = mx ± $\sqrt{a^{2} m^{2} - b^{2}}$

$∴$ point of contact $(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}); (\frac{- 9}{2 \sqrt{2}}, \frac{- 1}{\sqrt{2}})$