y=x^2+2e^x+2 at (0,4). (find the equation of tangent and normal to the curve at the point on it)
Answers
EXPLANATION.
Equation.
⇒ y = x² + 2eˣ + 2 at (0,4).
As we know that,
We can differentiate w.r.t x, we get.
⇒ dy/dx = d(x² + 2eˣ + 2)/dx.
⇒ dy/dx = 2x + 2eˣ + 0.
Put the value of x = 0 & y = 4 in equation, we get.
⇒ dy/dx = 2(0) + 2e⁽⁰⁾.
⇒ dy/dx = 2e⁽⁰⁾ = 2.
⇒ Slope of the tangent = 2.
As we know that,
Equation of tangent.
⇒ (y - y₁) = m(x - x₁).
Put the value in the equation, we get.
⇒ (y - 4) = 2(x - 0).
⇒ y - 4 = 2x.
⇒ 2x - y + 4 = 0.
Equation of normal.
⇒ (y - y₁) = -1/m(x - x₁).
Slope of normal = -1/m = -1/2.
Put the value in the equation, we get.
⇒ (y - 4) = -1/2(x - 0).
⇒ 2(y - 4) = -(x).
⇒ 2y - 8 = -x.
⇒ 2y + x - 8 = 0.
MORE INFORMATION.
(1) = The inclination of the tangent with x-axes = tan⁻¹(dy/dx).
(2) = Slope of tangent = (dy/dx) at (x₁, y₁).
(3) = Slope of normal = -(dx/dy) at (x₁, y₁).
Answer:
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