#### To determine

**To find: **Equation of tangent line and normal line to the given curve at given point.

#### Answer

Equation of tangent line:

y=-4x5+135

Equation of normal line:

y=5x4-32

#### Explanation

Differentiate the given curve and then plug the given point in it to get slope of tangent line. Then use point slope form of line to get equation of tangent line. Then calculate the negative reciprocal of slope of tangent line. It is slope of normal line. Again with this new slope find the equation of normal line using the same slope point form.

**1) Formula:**

i. Power rule combined with chain rule:

ddxf(x)n=n.fxn-1.f'(x)

ii. Sum rule:

ddxfx+gx=ddxfx+ddxg(x)

iii. Constant multiple rule:

ddxk.fx=k.ddxfx where k is constant.

iv.

ddxsinx=cosx

v. Constant function rule:

ddxk=0 where k is constant.

vi. Normal line and tangent lines are perpendicular to each other.

vii. Slopes of perpendicular lines are negative reciprocal of each other.

viii. Equation of line passing through point (a,f(a)) with slope m is, (y-f(a))=m(x-a)

**2) Given:**

x2+4xy+y2=13, (2 ,1)

**3) Calculation:**

Differentiate given equation with respect to x,

ddx(x2+4xy+y2)=ddx13

Using power rule, product rule and constant function rule,

2x+4.xddxy+y.ddxx+2y.dydx=0

Using power rule,

2x+4xdydx+4y+2y.dydx=0

x+2xdydx+2y+y.dydx=0

Taking dydx common,

x+(2x+y)dydx+2y=0

dydx=-2y-x(2x+y)

At (2 ,1)

f'(2)=-2*1-2(2*2+1)=-45

Equation of tangent line to the curve x2+4xy+y2=13 at (2,1) is

y-1=-45*x-2

y-1=-4x5+85

adding 1 on both side,

y=-4x5+85+1

y=-4x5+135

This is the required equation of tangent line.

To get equation of normal line, find the negative reciprocal of slope of tangent line.

So, slope of normal line = 54

Using point slope form to get normal line,

y-1=54*x-2

y-1=5x4-104

adding 1 on both side,

y=5x4-104+1

y=5x4-32

This is the required equation of normal line.

**Conclusion:**

Equation of tangent line:

y=-4x5+135

Equation of normal line:

y=5x4-32