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» finding slope of the tangent by differentiating the curve \(2x+4y\frac{dy}{dx}-4-6\frac{dy}{dx}=0\) \(\frac{dy}{dx}=\frac{4-2x}{4y-6}\) Finding y co – ordinate by substituting 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 normal is perpendicular to tangent so, m₁m₂ = – 1 m(normal) at x = 2 is \(\frac{1}{0}\), which is undefined 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»

finding slope of the tangent by differentiating the curve

\(2x+4y\frac{dy}{dx}-4-6\frac{dy}{dx}=0\)

\(\frac{dy}{dx}=\frac{4-2x}{4y-6}\)

Finding y co – ordinate by substituting 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

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at x = 2 is \(\frac{1}{0}\), which is undefined

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

x = 2