Find the equations of the tangent and the normal, to curve 16x² + 9y² = 145 at the point (x₁,y₁) where x₁ = 2 and y₁ > 0

Solution

16x² + 9y² = 145 is the curve and points (x₁,y₁) where

x₁ = 2 and y₁ >0

⇒ 16 (2)2 + 9

$\Rightarrow 16 (2)^{2} + 9 y_{1}^{2} = 145 \Rightarrow 9 y_{1}^{2} = 145 - 64 = 81 \Rightarrow y_{1}^{2} = 9 y_{1} = \pm 3 since y_{1} > 0 ∴ y_{1} = 3$

So, the required point is (2,3). Now 16x² + 9y² = 145, on differentiating w.r.t. x gives

$16 (2 x) + 18 y \frac{dy}{dx} = 0 \frac{dy}{dx} = - \frac{32 x}{18 y} = - \frac{16 x}{9 y} Slope of tangent at (2, 3) = {\frac{dy}{dx}}_{(2, 3)} = \frac{16 x 2}{9 x 3} = - \frac{32}{27} So, equation of tangent is y - y_{1} = m (x - x_{1}) y - 3 = \frac{- 32}{27} (x - 2) 27 y - 81 = 32 x + 64 \Rightarrow 323 x + 27 y = 145 is the equation of tangent slope of normal is \frac{- 1}{m} = \frac{- 1}{\frac{- 32}{27}} = \frac{27}{32} Equation of normal is y - 3 = \frac{27}{32} (x - 2) \Rightarrow 32 y - 96 = 27 x - 54 \Rightarrow 27 x - 32 y = 54 - 96 = - 42 \Rightarrow 27 x - 32 + 42 =$

For Daily Free Study Material Join wiredfaculty

Application Of Derivatives

Find the equations of the tangent and the normal, to curve 16x² + 9y² = 145 at the point (x₁,y₁) where x₁ = 2 and y₁ > 0

Some More Questions From Application of Derivatives Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction

For Daily Free Study Material Join wiredfaculty

Application Of Derivatives

Find the equations of the tangent and the normal, to curve 16x2 + 9y2 = 145 at the point (x1,y1) where x1 = 2 and y1 > 0

Some More Questions From Application of Derivatives Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Find the equations of the tangent and the normal, to curve 16x² + 9y² = 145 at the point (x₁,y₁) where x₁ = 2 and y₁ > 0