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प्रश्न
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
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उत्तर
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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