Question

y=x^3-x*y^2-4x^2-xy+5x+3y+2=0 find the equation of tangent and normal to the curve at point (1,-1)​

Answer 1

Answer:

Equation of curve is y = 3x2 – x + 1    ………… (1)

Differentiating equation (1) w.r.t. x

(dy/dx) = 6x – 1

∴  (dy/dx)at P(1, 3) = 6(1) – 1 = 6 – 1 = 5

The slope of the tangent at P(1, 3) is 5 and that of normal is – 1/5.

Equation of tangent:

Its slope = m = 5

It passes through P(1, 3) ≡ P(x1, y1)

By slope point form

y – y1 = m(x – x1)

∴  y – 3 = 5(x – 1)

∴  y – 3 = 5x – 5

∴  5x – y – 2 = 0

Equation of normal:

Its slope = m = – 1/5

It passes through P(1, 3) ≡ P(x1, y1)

By slope point form

y – y1 = m(x – x1)

∴  y – 3 = (- 1/5)(x – 1)

∴  5y – 15 = -x + 1

∴  x + 5y -15 – 1 = 0

∴  x + 5y – 16 = 0

Ans: Equation of tangent is 5x – y – 2 = 0 and that of normal is x + 5y – 16 = 0

Step-by-step explanation:

Answer 2

Answer:

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