Question

Find the equations of tangent and normal to the curve y=x2-2x+1 at the point where it cuts the x-axis.​

Answer 1

Answer:

Given, equation of curve is

y=

(x−2)(x−3)

(x−7)

....(1)

To find the intersection of given curve with X-axis, Put y=0 in equation 1, we get,

0=

(x−2)(x−3)

x−7

x−7=0⟹x=7

Thus, the curve cut the X-axis at (7,0).

Now, on differentiating equation of curve w.r.t. x, we get,

dx

dy

=

[(x−2)(x−3)]

2

(x−2)(x−3).1−(x−7)[(x−2).1+(x−3).1]

=

[(x−2)(x−3)]

2

(x−2)(x−3)−(x−7)(2x−5)

=

[(x−2)(x−3)]

2

(x−2)(x−3)[1−

(x−2)(x−3)

(x−7)

(2x−5)]

=

(x−2)(x−3)

1−y(2x−5)

(

dx

dy

)

(7,0)

=

20

1

Thus, slope of tangent =

20

1

Slope of normal = −20

Hence, the equation of tangent at (7,0) is

y−0=

20

1

(x−7)⟹20y−x+7=0

And, the equation of normal at (7,0) is

y−0=−20(x−7)⟹20x+y−140=0