Question

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

Answer 1

Given curve 16x²+9y²=145     ... (i)

Putting x₁=2 in (i), we get

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

`64 + 9y^2 = 145`

`9y^2 = 81 `

`y^2 = 9`

`y = +-3`

Given that y > 0 so y = 3

`y^2 = (145-16x^2)/9`        From i

Differentiating above both sides with respect to x

2y (dy)/(dx)

` = 1/9 (0- 32x)`

`(dy)/dx = (-32x_1)/(18y)`

`((dy)/(dx))_((2 "," 3)) = (-64)/54 = (-32)/27`

`:. (dy/dx)_((2","3)) = -32/27`

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

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

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

`=> 27y - 81 = -32x + 54`

`=> 32x + 27y - 135 = 0`

Now, the normal to the curve at (x_1, y₁) will be perpendicular to the tangent to the curve at (x₁, y₁)

Let normal to the curve have the slope m₁

Then `m_1 xx((-32)/27) = -1`

`:.m_1 = (27/32)`

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

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

`=> 32y−96=27x−54`

`=> 27x - 32y - 42 = 0`