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

1 Answer | Add Yours

Top Answer

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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)`

We’ve answered 315,839 questions. We can answer yours, too.

Ask a question