$\displaystyle f(x)=x^2-4x+n$ and the point which is not on the parabola (1,0) are given.If the tangent lines which drawn from (1,0) to parabola are perpendicular to each other, find n ? Thx for help
the slope of the tangents to the parabola is given by:
$\displaystyle \frac{d}
{{dx}}f(x) = 2x - 4$
let us denote the x coordinate (where it touches the parabola) of the first line as:$\displaystyle x_1$ and that of the second line as $\displaystyle x_2$, now we're told that the two tangents are perpendicular thus:
$\displaystyle \left( {2x_1 - 4} \right)\left( {2x_2 - 4} \right) = - 1$
the equations of the two lines are:
$\displaystyle
\left\{ \begin{gathered}
y = \left( {2x_1 - 4} \right)\left( {x - x_1 } \right) + x_1 ^2 - 4x_1 + n \hfill \\
y = \left( {2x_2 - 4} \right)\left( {x - x_2 } \right) + x_2 ^2 - 4x_2 + n \hfill \\
\end{gathered} \right.
$
we know that the two lines intersect at (1,0), thus:
$\displaystyle
\left( {2x_1 - 4} \right)\left( {1 - x_1 } \right) + x_1 ^2 - 4x_1 + n = \left( {2x_2 - 4} \right)\left( {1 - x_2 } \right) + x_2 ^2 - 4x_2 + n
$
so we've got two equation in two unknowns $\displaystyle x_1, x_2$, find them and then finding n is trivial...