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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=−x20.

(1)

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

k=−2x0.

(2)

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

y0=kx0+m.

(3)

In addition to this, the tangent should also pass through the point (1,1),

1=k1+m.

(4)

Equations (1)-(4) constitute a system of equations 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=−2x0 and substituting this into equation (4) gives

\displaystyle 1 = -2x_0 + m\quad\Leftrightarrow\quad m = 2x_0+1\,\textrm{.}

With k and m expressed in terms of \displaystyle x_0 and \displaystyle y_0, (3) becomes an equation that is expressed completely in terms of \displaystyle x_0 and \displaystyle y_0,

\displaystyle y_0 = -2x_0^2 + 2x_0 + 1\,\textrm{.}

(3')

This equation, together with (1), is a system of equations in \displaystyle x_0 and \displaystyle y_0,

\displaystyle \left\{\begin{align} y_{0} &= -x_0^{2}\,,\\[5pt] y_{0} &= -2x_0^2 + 2x_0 + 1\,\textrm{.} \end{align}\right.

Substituting equation (1) into (3') gives us an equation in \displaystyle x_0,

\displaystyle -x_0^2 = -2x_0^2 + 2x_0 + 1\,,

i.e.

\displaystyle x_0^2 - 2x_0 - 1 = 0\,\textrm{.}

This quadratic equation has solutions

\displaystyle x_0 = 1-\sqrt{2}\qquad\text{and}\qquad x_0 = 1+\sqrt{2}\,\textrm{.}

Equation (1) gives the corresponding y-values,

\displaystyle y_0 = -3+2\sqrt{2}\qquad\text{and}\qquad y_0 = -3-2\sqrt{2}\,\textrm{.}

Thus, the answers are the points \displaystyle (1-\sqrt{2},-3+2\sqrt{2}) and \displaystyle (1+\sqrt{2},-3-2\sqrt{2})\,.