Lösung 1.1:4

Aus Online Mathematik Brückenkurs 2

If we write the equation of the tangent as

y=kx+m

we know that the tangent's gradient k is equal to the derivative of y=x2 at the point x=1, and since

y=2x, so

\displaystyle k={y}'\left( 1 \right)=2\centerdot 1=2

We can determine the constant \displaystyle m\text{ } with the condition that the tangent should go through the grazing point \displaystyle \left( 1 \right.,\left. 1 \right), i.e. the point \displaystyle \left( 1 \right.,\left. 1 \right) should satisfy the equation of the tangent

\displaystyle 1=2\centerdot 1+m

which gives that \displaystyle m=-\text{1}

The normal to the curve \displaystyle y=x^{\text{2}}\text{ } at the point \displaystyle \left( 1 \right.,\left. 1 \right) is the straight line which is perpendicular to the tangent at the same point.

Because two straight lines which are perpendicular to each other have gradients which satisfy \displaystyle k_{1}\centerdot k_{2}=-1, the normal must have a gradient which is equal to

\displaystyle -\frac{1}{k}=-\frac{1}{2}

The equation of the normal can therefore be written as

\displaystyle y=-\frac{1}{2}x+n

where \displaystyle n is some constant.

Since the normal must pass through the line \displaystyle \left( 1 \right.,\left. 1 \right), we can determine the constant \displaystyle n if we substitute the point into the equation of the normal,

\displaystyle 1=-\frac{1}{2}\centerdot +n

and this gives \displaystyle n=\frac{3}{2}.