Question

Show that the tangent to the curve 25x^5+5x^4-45x^3-5x^2+2x+6y-24=0 at the point (-1,1) is also a normal at two points of the curve.

Answer 1

whewww!

125x^4 + 20x^3 - 135x^2 - 10x + 2 + 6dy/dx = 0
at (-1,1)
125 - 20 - 135 + 10 - 2 + 6dy/dx = 0

6dy/dx = 22, so the slope of the tangent at (-1,1) is 22/6 = 11/3
equation of tangent:
1= (11/3)(-1) + b
b = 1 + 11/3 = 14/3

y = (11/3)x + 14/3 = (11x+14)/3

sub that into the original equation
25x^5+5x^4-45x^3-5x^2+2x+6y-24=0
25x^5+5x^4-45x^3-5x^2+2x+6(11x+14)/3-24=0
25x^5+5x^4-45x^3-5x^2 + 24x + 4=0

after dividing out the factor (x+1) we are left with
25x^4 - 20x^3 - 25x^2 +20x+4=0

tried to solve this, even had Wolfram page try it
http://www.wolframalpha.com/input/?i=25x%5E4+-+20x%5E3+-+25x%5E2+%2B20x%2B4%3D0

it gave me x = appr -.9507 and appr -.169872

If we sub those values into the derivative equation we should get -3/11 or -.2727..
I did not get that.

Can't find my error
OR,
those two points don't exist
(you might print out my solution and go through it to see if find any errors)

Answer 2

-6y = 25x^5+5x^4-45x^3-5x^2+2x+24
y' = -1/6 (125x^4 + 20x^3 - 135x^2 - 10x + 2)
y'(-1) = 3
normal has slope -1/3

Go to http://rechneronline.de/function-graphs/

and plot

-1/6 * (25x^5+5x^4-45x^3-5x^2+2x-24)
and its derivative
as well as
3x+4

The line is not normal to the curve anywhere. Typo somewhere here?

Answer 3

When plotting the graphs, set
x: -2 to 2
y: -10 to 10

that will show things quite clearly.