Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. x=t√,y=t2−2t;t=4

Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. x=√t, y=t^2−2t; t=4

Answer 1

[Wan Ji]

syms t x0(t) y0(t) x y
x0 = sqrt(t); % parametric equation for x
y0 = t^2-2*t; % parametric equation for y
dx = diff(x0); % dx/dt
dy = diff(y0); % dy/dt
eq = subs(dy,t,4)*(x-subs(x0,4)) - subs(dx,t,4)*(y-subs(y0,4)) % this is the line eqaution eq=0

The answer then becomes

eq =
 
6*x - y/4 - 10

So 6*x - y/4 - 10 = 0 is the equation of the tangent to the curve at t = 4.

You can also solve it to extract y (in that case, the slope should not be inf)

y = solve(eq, y)
 
y =
 
24*x - 40

[Kevin Thongkham]

Thank you so much! Amazing!

Answer 2

[Kevin Thongkham]

Identify the type of conic section whose equation is given and find the vertices and foci. y^2−2=x^2−2x