Question

Show that the line 3x – 4y + 10 = 0 is tangent till the hyperbola x² – 4y² = 20. Also find the point of contact

Answer 1

Given equation of the hyperbola is x² – 4y² = 20.

∴ `x^2/20 - y^2/5` = 1

Comparing this equation with `x^2/"a"^2 - y^2/"b"^2` = 1, we get,

a² = 20 and b² = 5

Given equation of line is 3x – 4y + 10 = 0

∴ y = `(3x)/4 + 5/2`

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

m = `3/4` and c = `5/2`

For the line y = mx + c to be a tangent to the hyperbola `x^2/"a"^2 - y^2/"b"^2` = 1, we must have 

c² = a²m² – b²

c² = `(5/2)^2 = 25/4`

∴ a²m² – b² = `20(3/4)^2 - 5`

= `20(9/16) - 5`

= `45/4-5`

= `25/4`

= c²

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

= `((-"a"^2"m")/"c", (-"b"^2)/"c")`

= `((-20(3/4))/((5/2)), (-5)/((5/2)))`

= (– 6, – 2)