Advertisement Remove all ads

Find the Equation of a Tangent and the Normal to the Curve Y = ( X − 7 ) ( X − 2 ) ( X − 3 ) at the Point Where It Cuts the X-axis - Mathematics

Question

Sum

Find the equation of a tangent and the normal to the curve `"y" = (("x" - 7))/(("x"-2)("x"-3)` at the point where it cuts the x-axis

Solution

Equation of the curve is given by

`"y" = (("x" - 7))/(("x"-2)("x"-3)`       ....(i)

The given curve cuts the x-axis so, y = 0

Putting the value y=0 in (i) we get x=7.

Differentiate (i) w.r.t. x, then

`(d"y")/(d"x") = (("x"-2)·("x"-3)-("x"-7)·(2"x"-5))/(("x"-2)^2 .("x"-3)^2`

The slope of the tangent to (i) at point (7,0) is given by

`"m"_t` = \[\left.\frac{dy}{dx}\right|_{(7,0)}\] = `(20)/(400) = (1)/(20)`

Equation of the tangent to (i) at point (7,0) is given by

`("y"-0) = (1)/(20) ("x"-7) ⇒ "x"-20"y"-7 = 0`

We know,

`"m"_t ·"m"_n = -1`

⇒ `"m"_n = (-1)/(1/20) = -20`

Therefore, slope of the normal to (i) at point (7,0) is given by

mn = − 20
Equation of the normal to (i) at point (7,0) is given by

Equation of the normal is (y - 0) = - 20( x - 7) ⇒ 20x + y - 140 = 0.

  Is there an error in this question or solution?
Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications
Login
Create free account


      Forgot password?
View in app×