Posted by **Avik** on Sunday, February 3, 2013 at 3:17pm.

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.

- Math -
**Reiny**, Sunday, February 3, 2013 at 5:03pm
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)

- Math -
**Steve**, Sunday, February 3, 2013 at 5:12pm
-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?

- Math - PS -
**Steve**, Sunday, February 3, 2013 at 5:43pm
When plotting the graphs, set

x: -2 to 2

y: -10 to 10

that will show things quite clearly.

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