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The tangent to the curve y(1+x^2) = 2 at the point (3,1/5) meets the curve again at Q....

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roshan-rox | Valedictorian

Posted June 29, 2013 at 4:16 AM via web

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The tangent to the curve y(1+x^2) = 2 at the point (3,1/5) meets the curve again at Q. Find Q.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted June 29, 2013 at 4:26 AM (Answer #1)

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The tangent to the curve y(1+x^2) = 2 at the point (3, 1/5) meets the curve again at Q.

For a curve y = f(x), the slope of the tangent at any point with x = a is f'(a).

y(1+x^2) = 2

=> `y = 2/(1+x^2)`

=> `dy/dx = (-4*x)/(x^4+2*x^2+1) `

At x = 3, `dy/dx` = -0.12

If the point Q has coordinates `(x, 2/(1+x^2))`

`(2/(1+x^2) - 1/5)/(x - 3) = -0.12`

=> `x = -4/3`

`y = 18/25`

The coordinates of point Q are `(-4/3, 18/25)`

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