Line tangent to a parabola

Question

The plane of equation x + y = 1 intersects the cone of equation z = 4 − √((x^2)+(y^2)) in a parabola. Verify that the point of coordinates (3/7, 4/7, 23/7)
is on that parabola and find the equation of the line tangent to the
parabola at the given point.
Hint: Parameterize the parabola by x = t, y =?, z =?.

 

Not sure at all.

 

Thank you.

Answer 1

Parametrization will work like this:

 

x=t

x+y=1 => y=1-x => y=1-t

z=4-√(x^2+y^2) ⇒ z=4-√(t^2+t^2-2t+1) or z=4-√(2t^2-2t+1)

 

So our parametrization for the curve is <t, 1-t, 4-√(2t^2-2t+1)>

 

Plug in 3/7 for t and verify that x=t=3/7, y=1-t=4/7 and z=4-√(2t^2-2t+1)=23/7

 

In order to find the equation of a tangent at that point, we need the point and the tangent vector at that point.

 

Tangent vector for the curve is the derivative of the vector function f(t) <t, 1-t, 4-√(2t^2-2t+1)>

 

The derivative of the above function is f'(t) = <1, -1, (1-2t)/(√(2t^2-2t+1)>

 

f'(3/7) = <1, -1, 1/5>

 

The parametric equation of the tangent is <3/7+t, 4/7-t,23/7+t/5>