Conics Tangent/Gradient Question (1 Viewer)

[cutemouse]

Hello, I'm having trouble with this question. The gradient that I get is undefined. How do I work around it? Thanks

Find the equation of the tangent and normal to 4x²+y²-16x-2y+13=0 at (3,1)

Thanks again!

 

[shady145]

that means the tangent is vertical and is given by the equation
x=3
the normal then is horizontal given by
y=1
i got the gradient to be undefined too.

 

[cutemouse]

Hmm okay... Why would the tangent be vertical if dy/dx is undefined?

 

[Trebla]

Hmm okay... Why would the tangent be vertical if dy/dx is undefined?

As the gradient gets steeper and steeper, dy/dx --> infinity (i.e. undefined)

 

[shaon0]

Hello, I'm having trouble with this question. The gradient that I get is undefined. How do I work around it? Thanks

Find the equation of the tangent and normal to 4x²+y²-16x-2y+13=0 at (3,1)

Thanks again!

d/dx(4x^2+y^2-16x-2y+13)=d/dx(0)
8x+2yy'-16-2y'=0
y'=-4(x-2)/(y-1)
Let x=3, y=1.
y'=-4(1)/0 ie. Vertical tangent. (Critical pt)

 

[shady145]

because undefined is also infinite.
so there is rise, but no run, therefore has to be vertical, parallel to the y axis.
take an example where there is just 2 pts.
use the gradient formula .
y1-y2/x1-x2
if x1-x2=0 then the points are above and below each other, and if u get 0 on bottom of fraction then it is undefined, therefore is vertical.

find the gradient for the normal and u will get 0.
because, there is no rise and only run. so parallel to the x axis.