A curve is given parametrically by the equations, $x = (1+t)^2,y = (1-t)^2$.

Question

A curve is given parametrically by the equations, $x = (1+t)^2, y = (1-t)^2$.

Find the equation of the tangent to the curve at the point where $x = y$.

Answer 1

Let $x=y$. Then $$(1+t)^2=(1-t)^2.$$ Thus，$$t=0.$$ This shows the tangent point is $(1,1).$

Hence, let the slope of the tangent line be $k$. Then,$$k=\frac{{\rm d} y}{{\rm d}x}\bigg|_{x=1}=\frac{\dfrac{{\rm d} y}{{\rm d}t}}{\dfrac{{\rm d} x}{{\rm d}t}}=\frac{2t+2}{2t-2}\bigg|_{t=0}=-1.$$

Therefore, the equation of the tangent line is that $$y-1=-1(x-1),$$namely,$$y=-x+2.$$

Answer 2

x = y when:

(1 + t)^2 = (1 - t)^2

Therefore t^2 + 2t + 1 = t^2 - 2t + 1

Therefore 2t = -2t

Which only holds true when t = 0

When t = 0, x and y both equal 1 i.e. (1, 1)

Graphing x=(1+t)^2, y=(1−t)^2

From visually examining this graph, it should be clear that the tangent to the curve at (1,1) has the equation x + y = 2