find the equationss of both tangenr lines to the curve defined by y=(1/x^2), x cannot equal 0 from the external point P(0,1).
I tried but i got one equation. Can you please verify if it is: y=-0.38x-1.
Thanks
You are close, but are not correct.
Your slope is correct, but you have only one tangent.
Secondly since the tangent must go through (0,1) the y-intercept should be 1
you have y = .38x - 1
Here is how I would do this:
Let the point of contact of the tangent(s) be (a,b)
y = 1/x^2
dy/dx = -2/x^3
so at (a,b) the slope of the tangent is -2/a^3
but the slope can also be found the "old fashioned grade 9 method",
slope of tangent = (b-1)/(b-0) = (b-1)/a
but we know b = 1/a^2, since (a,b) is on the curve.
then
-2/a^3 = (b-1)/a
-2/a^3 = (1/a^2 - 1)a
-2/a^2 = 1/a^2 - 1
-2 = 1 - a^2
a^2 = 3
a = ± √3, and b = 1/3
so there are two points of contact,
(√3,1/3) and (-√3,1/3)
also the slope of the tangent is
-2/(√3)^3 with a y-intercept of 1
so the two tangents are
y = (-2/(3√3))x + 1 and y = (2/(3√3))x + 1
btw, 2/(3√3) = .385 which is the slope you had.