Question

Show that the line 2x – y = 4 touches the hyperbola 4x² – 3y² = 24. Find the point of contact.

Answer 1

Given equation of die hyperbola is 4x² – 3y² = 24.  

∴ \(\frac {x^2}{16} - \frac {y^2}{8} = 1\)

 Comparing this equation with \(\frac {x^2}{a^2} - \frac {y^2}{b^2} = 1\)

we get 

a = 6 and b = 8 

Given equation of line is 2x – y = 4 

∴ y = 2x – 4 

Comparing this equation with y = mx + c, we get 

m = 2 and c = -4 

For the line y = mx + c to be a tangent to the hyperbola

\(\frac {x^2}{a^2} - \frac {y^2}{b^2} = 1\), we must have

c² = a² m² – b²

c² = (-4)² = 16 

a² m² – b² = 6(2)² – 8 = 24 – 8 = 16 

∴ The given line is a tangent to the given hyperbola and point of contact

= (3, 2)