Question

find the equation of tangent and normal, to the curve 16x^2+9y^2=145 at the point (x1,y1) , where x1= 2 and y1 > 0.

Answer 1

Given:

The curve 16x^2+9y^2=145 at the point (x1, y1) , where x1= 2 and y1 > 0.

To find:

Find the equation of tangent and normal to the curve 16x^2+9y^2=145 at the point (x1, y1).

Solution:

From given, we have the equation of curve 16x^2 + 9y^2 = 145

given: x1= 2 and y1 > 0.

substituting the value of x1 in given equation of curve, we have,

16(2)^2 + 9y^2 = 145

64 + 9y^2 = 145

9y^2 = 81

y = ± 3.

Given, y > 0 ⇒ y = +3

y^2 = (145 - 16x^2) /9

differentiating the above function with respect to x, we get,

2y dy/dx = 1/9 (0 - 32x)

dy/dx = - 32x/18y

(dy/dx)(2, 3) = - 32(2)/18(3) = - 32/27

The slope of tangent,

m = - 32/27

The equation of tangent at (2, 3) is,

y - 3 = (dy/dx)(2, 3) (x - 2)

y - 3 = (- 32/27) (x - 2)

solving the above equation, we get,

32x + 27y  - 135 = 0

The slope of normal is,

m1 × ( -32/27) = -1

m1 = 27/32

The equation of normal at (2, 3) is,

y - 3 = (27/32) (x - 2)

solving the above equation, we get,

27x - 32y  - 42 = 0