Question

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

Answer 1

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

m_n = − 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.