Processing Math: Done

Aus Online Mathematik Brückenkurs 2

Suppose that the tangent touches the curve at the point x0y0 . That point must, first and foremost, lie on the curve and therefore satisfy the equation of the curve, i.e.

y0=−x201 

If we now write the equation of the tangent as y=kx+m, the gradient of the tangent, k, is given by the value of the curve's derivative, y=−2x, at x=x0,

k=−2x02 

The condition that the tangent goes through the point x0y0  gives us that

y0=kx0+m3 

In addition to this, the tangent should also pass through the point 11 ,

1=k1+m4 

Equations (1)-(4) constitute an equation system in the unknowns x0 y0 k and m.

Because we are looking for x0 and y0, the first step is to try and eliminate k and m from the equations.

Equation (2) gives that k=−2 x0 and substituting this into equation (4) gives

1=−2x0+mm=2x0+1

With k and m expressed in terms of x0 and y0, (3) becomes an equation that is expressed completely in terms of x0 and y0,

y0=−2x20+2x0+13 

This equation, together with (1), is an equation system in x0 and y0

y0=−x20y0=−2x20+2x0+1 

Substituting equation (1) into (3') gives us an equation in x0,

−x20=−2x20+2x0+1

i.e.

x20−2x0−1=0

This second-degree equation has solutions

x0=1−2  and x0=1+2 

Equation (1) gives the corresponding y-values:

y0=−3+22  and y0=−3−22 

Thus, the answers are the points 1−2−3+22  and 1+2−3−22 .