# 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.

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