Question

x²/16 - y²/9=1 at(-5,9/4),Find the equation of the tangent to given curves at given point.

Answer 1

curve , x²/16 - y²/9 = 1

differentiate with respect to x,

2x/16 - 2y/9.dy/dx = 0

x/8 = 2y/9. dy/dx

dy/dx = 9x/16y

at, (-5,9/4) , slope of tangent of curve = $\frac{dy}{dx}|_{(-5,9/4)}$

= 9 × -5/(16 × 9/4)

= -45/36
= -5/4

now, equation of tangent :

passing through (-5, 9/4) and slope of tangent is -5/4

so, (y - 9/4) = -5/4(x + 5)

=> 4(y - 9/4) + 5(x + 5) = 0

=> 4y - 9 + 5x + 25 = 0

=> 5x + 4y + 16 = 0

hence, equation of tangent is 5x + 4y + 16

Answer 2

Dear Student:

Given: x²/16 - y²/9=1 at(-5,9/4)

$x_{0} =-5,y_{0} =9/4$

$m=\frac{dy}{dx} =slope of tangent$

See the attachment: