If the slope of the tangent to the curve x^2-2xy+4=0 at a point on it is -3/2, then find tangent and normal at that point (Applications of Differentiation) Can anyone help me with this?

Question

shubhamsrg · Answer

Whats your attempt/progress ?

anonymous · Answer

I found out the Y derivative of the curve and equated it to -3/2 and I got another equation in x and y terms... I don't know what to do after that....

shubhamsrg · Answer

One eqn you'll get after differentiating, 1 you already have. You know the slope. You just need a point, solve both eqns to get the req point.

anonymous · Answer

but one eqn is linear and the other quadratic hmm shou

anonymous · Answer

should i factorize the given curve eqn

anonymous · Answer

oopa no that's not possible

shubhamsrg · Answer

2x - 2y - 2xdy/dx = 0

x-y -3x/2 = 0
y = -x/2

IS that what you get ?

shubhamsrg · Answer

oh wait, correction :

it should be 
5x=2y

anonymous · Answer

I AM SOO SORRY THE Question is this...... $$x ^{2}-2xy+4y=0$$

shubhamsrg · Answer

hmm, so what eqns you get ?

anonymous · Answer

After diff and euating it to -3/2.... this is what I got.... $$5x-2y=6$$

shubhamsrg · Answer

okay. seems right.

shubhamsrg · Answer

original eqn is
x^2 - 2xy + 4y =0

its easier if you substitute for y from 1st eqn.

shubhamsrg · Answer

In case you are following ?

anonymous · Answer

I tried doing that but after substituting for Y the equation we get cannot be factorized directly... x^2+x-3=0

shubhamsrg · Answer

I am getting a solution, you have probably done mistake in substituing, tey again.

shubhamsrg · Answer

substituting* 
try**

shubhamsrg · Answer

problems ?

anonymous · Answer

hmm no am getting the same equation after substituting...

shubhamsrg · Answer

5x-2y =6
y = (5x-6)/2

anonymous · Answer

yes i substituted this in the original equation

shubhamsrg · Answer

x^2 - 2xy + 4y =0

x^2 - 2x(5x-6)/2 + 4(5x-6)/2 = 0

x^2 - x(5x-6) + 2(5x-6) = 0

anonymous · Answer

it should be 2x^2 in the last line

shubhamsrg · Answer

eh ?

anonymous · Answer

yea should take 2 as LCM and then send it to the other side, you forgot to multiply 2 to the first term x^2

shubhamsrg · Answer

I did not forget ,I cancelled out 2, one from "2"xy , other from "4"y

shubhamsrg · Answer

SO final eqn which you should be getting is
-4x^2 + 16x - 12 = 0
or
x^2 - 4x + 3 =0

shubhamsrg · Answer

And am sorry for my late replies, am generally not slow, just busy somewhere else also.

anonymous · Answer

ohh yeaaa right right :) thank you so much :) I got the points thanks you.... !!

shubhamsrg · Answer

Glad to help ^_^