Question

Q−Find the equation of the tangent line to the graph of x3+y3-3xy=0 at the point 32,32.​

Answer 1

The tangent is a straight line so it will be of the form

y

=

m

x

+

c

We can get

m

by finding the 1st derivative

d

y

d

x

as this is the gradient of the line.

We can then get

c

, the intercept, by using the values of

x

and

y

which are given.

To find the 1st derivative we can use implicit differentiation:

x

2

+

x

y

+

y

2

=

3

D

(

x

2

+

x

y

+

y

2

)

=

D

(

3

)

Using The Product Rule and The Chain rule gives:

2

x

+

x

y

'

+

y

+

2

y

y

'

=

0

∴

2

x

+

y

'

(

x

+

2

y

)

+

y

=

0

∴

y

'

(

x

+

2

y

)

=

−

2

x

−

y

∴

y

'

=

−

2

x

−

y

x

+

2

y

We are told

x

=

1

and

y

=

1

∴

y

'

=

−

2

−

1

1

+

2

=

−

3

3

=

−

1

This corresponds to the gradient

m

.

The tangent line is of the form

y

=

m

x

+

c

Putting in the values:

1

=

−

1

×

1

+

c

∴

c

=

2

So the equation of the tangent line becomes:

y

=

−

x

+

2

Or

y

=

2

−

x

The situation looks like this:

graph{(x^2+xy+y^2-3)(2-x-y)=0 [-10, 10, -5, 5]}

I hope this helps you im not sure if will but I hope you understand it