find the equations of tangent and normal to the given curve at the indicated point:
x^2/a^2 +y^2/b^2 =1 at (acos theta,bsin theta).
Answers
EXPLANATION.
Equation of tangent and equation of normal to the giving curve to the.
⇒ x²/a² + y²/b² = 1. at (a cosθ, b sinθ).
As we know that,
⇒ x²/a² + y²/b² = 1.
Differentiate w.r.t x, we get.
⇒ 2x/a² + 2y/b². (dy/dx) = 0.
⇒ 2y/b². (dy/dx) = - 2x/a².
⇒ dy/dx = (-2x/a²)/(2y/b²).
⇒ dy/dx = (-2x/a²) x (b²/2y).
⇒ dy/dx = -xb²/ya².
Put the value of (a cosθ, b sinθ) in the equation, we get.
⇒ dy/dx = -(a cosθ) x b²/(b sinθ) x a².
⇒ dy/dx = (- bcosθ)/(a sinθ).
As we know that,
Formula of equation of tangent.
⇒ (y - y₁) = m(x - x₁).
Put the value in the equation, we get.
⇒ (y - b sinθ) = (- bcosθ)/(a sinθ) (x - a cosθ).
⇒ (a sinθ)(y - b sinθ) = (- bcosθ)(x - a cosθ).
⇒ (a sinθ)y - absin²θ = (- bcosθ)x + abcos²θ.
⇒ ay sinθ + bx cosθ = abcos²θ + absin²θ.
⇒ ay sinθ + bx cosθ = ab.
⇒ bx cosθ + ay sinθ = ab.
⇒ (bx cosθ)/ab + (ay sinθ)/ab = 1.
⇒ x/a cosθ + y/b sinθ = 1.
As we know that,
Formula of equation of normal.
⇒ (y - y₁) = -1/m(x - x₁).
⇒ (y - b sinθ) = (a sinθ)/(b cosθ) (x - a cosθ).
⇒ (b cosθ)(y - b sinθ) = (a sinθ)(x - a cosθ).
⇒ by cosθ - b²sinθ.cosθ = ax sinθ - a²sinθ.cosθ.
⇒ ax sinθ - by cosθ = (a² - b²) sinθ.cosθ.
⇒ (ax sinθ)/(sinθ.cosθ) - (by cosθ)/(sinθ.cosθ) = (a² - b²).
⇒ (ax)/cosθ - (by)/sinθ) = (a² - b²).
⇒ ax secθ - by cosecθ = (a² - b²).
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