Question

(ii) Find the equation of normal to the curve y=x + x2-11+15 where normal is parallel to the
х 14 y + 4 = 0.
line x - 3y + 1 = 0.
(P.B. 2016)
1 where normal is parallel to​

Answer 1

Answer:

Given equation of curve is

y=x

3

+2x+6 .....(1)

Differentiating w.r.t x, we get

dx

dy

=3x

2

+2

Let P(x

1

,y

1

) be any point on the curve.

Slope of the tangent to the given curve at P(x

1

,y

1

) is

(

dx

dy

)

(x

1

,y

1

)

=3x

1

2

+2

Slope of the normal to the given curve at point P(x

1

,y

1

) is

Slope of the tangent at the point(x

1

,y

1

)

−1

=

3x

2

+2

−1

The equation of the given line is x+14y+4=0 which can be written as

y=−

14

1

x−

14

4

(which is of the form y=mx+c)

So, Slope of this line = −

14

1

Since, the normal is parallel to this line.

So, slope of normal = slope of the given line.

∴

3x

1

2

+2

−1

=

14

−1

⇒3x

1

2

+2=14

⇒3x

1

2

=12

x

1

2

=4

x

1

=±2

Since, P(x

1

,y

1

) lies on the curve

y

1

=x

1

3

+2x

1

+6 (by (1))

When x

1

=2,

⇒y

1

=8+4+6=18

When x

1

=−2,

⇒y

1

=−8−4+6=−6

Therefore, there are two normals to the given curve with slope

14

−1

and passing through the points (2,18) and (−2,−6).

Thus, the equation of the normal through (2,18) is given by,

y−18=

14

−1

(x−2)

⇒14y−252=−x+2

⇒x+14y−254=0

And, the equation of the normal through (−2,−6) is given by,

y−(−6)=

14

−1

[x−(−2)]

⇒y+6=

14

−1

(x+2)

⇒14y+84=−x−2

⇒x+14y+86=0

Hence, the equations of the normals to the given curve (which are parallel to the given line) arex+14y−254=0 and x+14y+86=0.