Find the equation of the normal to the curve x2 + 2y2 – 4x – 6y

1.	Find the equation of the normal to the curve x2 + 2y2 – 4x – 6y + 8 = 0 at the point whose abscissa is 2.
Answer» Given as x² + 2y² - 4x - 6y + 8 = 0 Differentiate with respect to x, to get the slope of tangent 2x + (4y(dy/dx)) - 4 - (6dy/dx) = 0 dy/dx = (4 - 2x)/(4y - 6) Finding the y co–ordinate by substitute x in the given curve 2y² – 6y + 4 = 0 y² – 3y + 2 = 0 y = 2 or y = 1 m(tangent) at x = 2 is 0 The normal is perpendicular to tangent therefore, m₁m₂ = – 1 m(normal) at x = 2 is 1/0, which is undefined The equation of normal is given by y – y₁ = m(normal)(x – x₁) x = 2

Find the equation of the normal to the curve x2 + 2y2 – 4x – 6y + 8 = 0 at the point whose abscissa is 2.

Answer»

Given as x² + 2y² - 4x - 6y + 8 = 0

Differentiate with respect to x, to get the slope of tangent

2x + (4y(dy/dx)) - 4 - (6dy/dx) = 0

dy/dx = (4 - 2x)/(4y - 6)

Finding the y co–ordinate by substitute x in the given curve

2y² – 6y + 4 = 0

y² – 3y + 2 = 0

y = 2 or y = 1

m(tangent) at x = 2 is 0

The normal is perpendicular to tangent therefore, m₁m₂ = – 1

m(normal) at x = 2 is 1/0, which is undefined

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

x = 2