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