equation of tangents and normals markscheme

7/27/2019 Equation of Tangents and Normals Markscheme

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EQUATIONS OF TANGENTS AND NORMALS MARKSCHEME

1.2

1

2

1

2

1

ayx =+

0d

d

2

1

2

12

1

2

1

=+

x

yyx

M1

x

y

y

x

x

y==

2

1

2

1

d

d

A1

Note: Accept2

1

2

1

1d

d

x

a

x

y=

from makingy the subject of the equation,

and all correct subsequent working

therefore the gradient at the point P is given by

p

q

x

y=

d

d

A1

equation of tangent isyq =

)( pxp

q

M1

(y =

pqqxp

q++

)

x-intercept:y = 0, n =

ppqpq

pq+=+

A1

y-intercept:x = 0, m =qpq +

A1

n+m =ppqppq +++

M1

=qppq ++2

=2)( qp +

A1

= a AG[8]

2. EITHER

differentiating implicitly:

1 ey

xey x

y

x

y y

d

de

d

d+

= 1 M1A1

at the point (c, ln c)

1d

d

d

d11=+

x

yc

x

y

cc

c M1

cx

y 1

d

d=

(c 1) (A1)


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OR

reasonable attempt to make expression explicit (M1)

xey

+ ey

= 1 +x

x + e2y

= ey

(1 +x)

e2y

ey

(1 +x) +x = 0

(ey 1)(ey x) = 0 (A1)ey

= 1, ey

=x

y = 0,y = lnx A1

Note: Do not penalize ify = 0 not stated.

2

1

d

d=

x

y

gradient of tangent = c

1

A1

Note: If candidate starts withy = ln x with no justification,

award (M0)(A0)A1A1.

THEN

the equation of the normal is

y ln c =c(x c) M1

x = 0,y = c2+ 1

c2

+ 1 ln c = c2

(A1)

ln c = 1

c = e A1[7]

3. x3y

3xy = 0

3x2y

3+ 3x

3y

2y y xy = 0 M1A1A1

Note: Award A1 for correctly differentiating each term.

x = 1,y = 1 3 + 3y 1 y = 0

2y =2

y = 1 (M1)A1

gradient of normal = 1 (A1)

equation of the normaly1 =x1 A1 N2 y =x

Note: Award A2R5 for correct answer and correct justification.[7]


3/7


4/7


5/7

[17]

6. xd

d

(arctan (x 1)) =( ) 211

1

+ x(or equivalent) A1

mN

= 2 and so mT

= 2

1

(R1)

Attempting to solve( ) 211

1

+ x= 2

1

(or equivalent) forx M1

x = 2 (asx > 0) A1

Substitutingx = 2 and 4

=y

to find c M1

c = 4 + 4

A1 N1

[6]

7. (a) (i)( ) 123 22 += kxxkxfk A1

( ) kxkxfk 262 =

A1

(ii) Settingf(x) = 0 M1

6k2x 2k= 0 x = k3

1

A1

+

=

kkk

kk

kf

3

1

3

1

3

1

3

123

2

M1

= k27

7

A1

Hence,Pk

is

kk 27

7,

3

1

(b) Equation of the straight line isxy

9

7=

A1

As this equation is independent ofk, allPk

lie on this straight line R1


6/7

(c) Gradient of tangent atPk:

( )3

21

3

12

3

13

3

12

2 =+

=

=

kk

kk

kfPf k

M1A1

As the gradient is independent ofk, the tangents are parallel. R1

kcc

kk 27

1

3

1

3

2

27

7=+=

(A1)

The equation isy = kx

27

1

3

2+

A1[13]

8. 5y2

+ 10xy x

y

d

d

4x = 0 A1A1A1

Note: Award A1A1 for correct differentiation of 5xy2.

A1 for correct differentiation of 2x2

and 18.

At the point (1, 2), 20 + 20 x

y

d

d

4 = 0

5

4

d

d=

x

y

(A1)

Gradient of normal = 4

5

A1

Equation of normaly 2 = 4

5

(x 1) M1

y = 4

8

4

5

4

5+x

y = 4

3

4

5+x

(4y = 5x + 3) A1[7]


7/7

9. Attempting to differentiate implicitly (M1)

3x2y + 2xy

2= 2

0d

d42

d

d36

22 =+++y

yxyy

x

yxxy

A1

Substitutingx = 1 andy = 2 (M1)

12 + 3

0

d

d88

d

d=+

x

y

x

y

A1

5

4

d

d4

d

d5 ==

x

y

x

y

A1

Gradient of normal is 4

5

A1 N3[6]

10. xey

=x2

+y2

xyyx

xyx yy

dd22

ddee +=+

M1A1A1A1A1

(1, 0) fits x

y

d

d1+

= 2 + 0

x

y

d

d

= 1 A1

Equation of tangent isy =x + c

(1, 0) fits c = 1 y =x 1 A1

[7]

equation of tangents and normals markscheme

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