how do i find the equations of the tangent lines to the curve 5X^2 - y^2 = 41 at the point (3,-2)..its urgent cuz i got a test in like 4 hours..

Question

how do i find the equations of the tangent lines to the curve 5X^2  - y^2 = 41 at the point (3,-2)..its urgent cuz i got a test in like 4 hours..

anonymous · Answer

@andesa , u know how to differenciate?

anonymous · Answer

let $$y_0 = -2 \text{ , } x_0 = 3$$
the equation of your line will be:
$$y - y_0 = m(x - x_0)$$
differentiating your curve eq. to find m:

$$5x^2 - y^2 = 41$$
$$10x - 2y\frac{dy}{dx} = 0$$
$$\frac{dy}{dx} = \frac{10x}{2y} = \frac{5x}{y}$$

at your point we have

$$m = \frac{dy}{dx} = \frac{5(3)}{-2} = \frac{-15}{2}$$

anonymous · Answer

this requires a bit of knowledge of implicit differentiation, are you ok with that?

anonymous · Answer

yes!..thanks a lot!!

anonymous · Answer

well, after differenciating, u will get:
10x-2y dy/dx=0
dy/dx= 10x/2y= 5x/y
slope of the tangent at any point (x,y)= 5x/y
u have to put the value of the given point, so, slope=5*3/(-2)=-15/2
now, equation of the tangent:
y=-15/2 x+c
as the line passes through (3,-2)..
-2=-15/2 *3+c
u will get the value of c, u will get the answer