Find the equation of the tangents to the curve 3x² – y² = 8, which passes through the point (4/3, 0). (CBSE 2013)

assume point (a, b) which lies on the given curve

finding the slope of the tangent by differentiating the curve

m(tangent) at (a,b) is

Since this tangent passes through , its slope can also be written as

Equating both the slopes as they are of the same tangent

b² = 3a² – 4a …(i)

Since points (a,b) lies on this curve

3a² – b² = 8 …(ii)

Solving (i) and (ii) we get

3a² – 8 = 3a² – 4a

a = 2

b = 2 or – 2

therefore points are (2,2) or (2, – 2)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 2 = 3(x – 2)

or

y + 2 = – 3(x – 3)