Algebra 02

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Intersection(s) of graph and line

Q: The graph of \(y=\frac 4x\) and a line through \([0,4]\) may intersect.
What slope of the line results in 2,1 or 0 intersections?
Where does the line touch the graph in just 1 point?

Graph:

Answer:

\( \begin{align} \frac 4x &=mx+4 &|\times x \\ 4 &=mx^2+4x &|-4 \\ mx^2 + 4x - 4 &=0 &|\div m \\ x^2 + \frac 4m x - \frac 4m & =0 \\ \left(x + \frac 2m\right)^2 - \left(\frac 2m\right)^2 - \frac 4m & =0 \\ \left(x + \frac 2m\right)^2 & = \frac{4}{m^2} + \frac 4m & =0 \\ x & = - \frac 2m \pm \sqrt{\frac{4}{m^2} + \frac 4m} \\ x & = - \frac 2m \pm \sqrt{\frac 4m \left(\frac 1m + 1\right)} \\ \end{align} \\ \)
\( \frac 4m \left(\frac 1m + 1 \right) = 0 \text{ results in 1 real solution exactly when } m = -1\\ \frac 4m \left(\frac 1m + 1\right) >0 \text{ results in 2 real solution exactly when } m < -1\\ \frac 4m \left(\frac 1m + 1\right) <0 \text{ results in 0 real solution exactly when } m > -1\\ \text{With }m=-1 \\ x=-\frac{2}{-1} \pm \sqrt{\frac{4}{-1}\left(\frac{1}{-1}+1\right)} \\ x=2 \pm \sqrt{(-4) \times \left(-1+1\right)} \\ x=2 \pm 0 \\ x=2 \\ y=\frac 4x =\frac 42 = 2\\ A=[2,2] \\ \)

Tangents of line on graph

Q: The graph of \(y=x^2-2x+2\) and a line through \([1,0]\) may intersect. What slope(s) of the line results in exactly 1 intersection?
Where do the lines touch the graph in just 1 point?

Graph:

Answer:

\( \begin{align} y & = x^2-2x+2, \text{ and } \\ y & = mx+k, \text{ with } y(+1)=0 \implies 0=m(+1) + k \implies k=-m \implies \\ y & = mx-m \\ \end{align} \\ \)
Graph and line intersect when:
\( \begin{align} mx-m &= x^2-2x+2 \\ x^2-mx-2x+m+2 &= 0 \\ x^2-(m+2)x+(m+2) &= 0 \\ \left(x-\frac{m+2}{2}\right)^2 - \left(\frac{m+2}{2}\right)^2 + (m+2) &= 0 \\ \left(x-\frac{m+2}{2}\right)^2 &= + \frac{(m+2)^2}{4} -m-2 \\ x-\frac{m+2}{2} &= \pm \sqrt{\frac{m^2 +4m+4}{4} -m-2} \\ x &= \frac{m+2}{2} \pm \sqrt{\frac{m^2 +4m+4}{4} +\frac{-4m-8}{4}}\\ x &= 1+\frac{m}{2} \pm \frac 12 \sqrt{m^2 +4m+4 -4m -8}\\ x &= 1+\frac{m}{2} \pm \frac 12 \sqrt{m^2 -4}\\ A=[2,2] \\ B=[0,2] \\ \end{align} \\ \)
This term has a single solution \(x\) if
\( \begin{align} m^2-4 &= 0\\ m^2 &= 4\\ m&=\pm\sqrt{4} \\ m &=\pm 2 \\ m_1 &=2 \\ m_2 &=-2 \\ \end{align} \\ y_1=f_1(x)=+2x-2 \implies x=1+1=2 \implies y(2)=2 \\ y_1=f_1(x)=-2x+2 \implies x=1+(-1)=0 \implies y(0)=2 \\ \)

Intersecting parabolas

Q: For which \(m\) do the graphs of \(y=x^2-2x+2\) and \(y=-x^2+m\) may just touch in one single point?

Graph:

Answer:

\( \begin{align} x^2-2x+2 &= -x^2+m &&| +x^2 -m \\ 2x^2-2x-m+2 &= 0 &&| \div 2 \\ x^2-x-\frac{m-2}{2} &&= 0 \\ \left(x-\frac 12\right)^2-\left(\frac 12\right)^2 -\frac{m-2}{2} &= 0 \\ \left(x-\frac 12\right)^2 &=\left(\frac 12\right)^2 +\frac{m-2}{2} \\ x &=+\frac 12 \pm \sqrt{\left(\frac 12\right)^2 +\frac{m-2}{2}} \\ \end{align} \)
There is exactly one solution if:
\( \begin{align} \sqrt{\left(\frac 12\right)^2 +\frac{m-2}{2}}&=0 &&\lvert \text{ square} \\ \left(\frac 12\right)^2 +\frac{m-2}{2}&=0 &&\lvert-\left(\frac 12\right)^2 \\ \frac{m-2}{2}&=-\frac 14 &&| \times 2 \\ m-2&=-\frac 12 &&| +2 \\ m&=1\frac 12 \\ \end{align} \)

Gradient on parabola point (algebraic solution)

Q: What is the function of the line touching the graphs of \(y=x^2\) at \(P=[1,1]\)?

Graph:

Answer:

\( \begin{align} y&=mx+b\\ \text{ Any line going through } [1,1] \text{ must fulfil: }\\ 1&=m\cdot 1+b &&| -m \\ b&=-m+1 && (*) \\ \text{ Therefore any line going through } [1,1] \text{ is: }\\ y&=mx-m+1 \\ \text{ Find } m \text{ for which is there only 1 solution to: } \\ x^2&=mx-m+1 &&|-mx \\ x^2-mx &=-m+1 &&| \text{ C.T.S.} \\ \left(x-\frac m2 \right)^2 - \left(\frac m2 \right)^2 &=-m+1 &&| +\left(\frac m2 \right)^2 \\ \left(x-\frac m2 \right)^2 &=-m+1 +\left(\frac m2 \right)^2 &&| \sqrt{} \\ x&=+\frac m2 \pm \sqrt{-m+1 +\frac {m^2}{4}} \\ \text{ There's exactly one solution if the root expression is } 0 \\ \frac{m^2}{4}-m+1&=0 &&| \times 4 \\ m^2-4m+4&=0 \\ (m-2)^2&=0 \\ m&=2 &&| \text{ use } (*)\\ b&=-m+1=-1 \\ y&=2x-1 \\ \end{align} \\ \)
A: The tangent function is \(y=2x-1\).

Points on Circle

The circle \(C\) passes through the points \(P\), \(Q\) and \(R\) with coordinates \((-2,-2)\), \((2,-4)\) and \((7,1)\) respectively.

Q (a): Find an equation of the perpendicular bisector of the points \(P\) and \(Q\).

Q (b): Find the coordinates of the centre of \(C\).

Q (c): Find an equation of \(C\).

Graph:

Answer a: perpendicular bisector of PQ:

For PQ with \( P=(-2,-2) \) and \( Q=(2,-4) \)
the perpendicular bisector is
\( y=mx+b \) with m being the negative inverse slope of the line PQ:
\(m=-\frac{1}{\frac{-4-(-2))}{2-(-2)}}=-\frac{1}{\frac{-2}{4}}=-\frac{1}{-0.5}=2\)
\( y=2x+b \)
with the midpoint of PQ as \( (0,-3) \) on that line, there is
\( -3=2(0)+b \\ b=-3 \)
and
\( \mathbf{y=2x-3} \)

Answer b: circle center (C)

Similarly, the halfway point between \( Q=(2,-4), R=(7,1) \) is
\( (\frac{2+7}{2},\frac{-4+1}{2})=(4.5,-1.5) \\ \) and the gradient is
\( -\frac{1}{\frac{7-2}{1-(-4)}}=-1 \\ \)
Therefore the line equation is
\( y=-x+b \)
with \( -1.5=-4.5+b \), therefore
\( b=3 \) , therefore the perpendicular bisector of QR is:
\( \mathbf{y=-x+3} \)
Those 2 lines intersect (at \( C \) ) where:
\( -x+3=2x-3 \\ 6=3x \\ x=2 \\ y=-x+3=-2+3=1 \)
Therefore the centre \( C \) of the circle is
\mathbf{C=(2,1)} \).

Answer c: circle equation:

The squared radius \( r^2 \) of the circle with \( C=(2,1) \) and \( R=(7,1) \):
\( r^2=(2-7)^2 + (1-1)^2=25 \)
Therefore the circle equation is:
\( \mathbf{(x-2)^2+(y-1)^2=25} \)

Parallel Tangent

The circle \(C\) has equation \(x^2+y^2-4x-6=0\) and the line \(l\) has equation \(y=3x-6\).

Q (a): Show that \(l\) passes through the centre of \(C\).

Q (b): Find an equation for each tangent to \(C\) that is parellel to \(l\).

Graph:

Answer a: \( l \) passing through centre of circle:

\( \begin{align} x^2+y^2-4x-6&=0 \\ (x-2)^2-4+y^2-6&=0 \\ (x-2)^2+y^2-10&=0 \\ (x-2)^2+y^2&=10 \\ \end{align} \)

The circle has its centre at \((2,0)\).

\( \begin{align} 3x-6&=0 \\ 3\cdot2-6&=0 \\ 0&=0 \checkmark \\ \end{align} \)

The line goes through \((2,0)\).

Answer b: equations of tangents parallel to \( l \)

A parallel line to \(l\) is \(y=3x+t\).

If \(y=3x+t\) intersects the circle, \((x,y)\) are identical for the circle and line equation at those points, i.e. \(y\) can be substituted.

\( \begin{align} x^2+(3x+t)^2-4x-6&=0 \\ x^2+9x^2+6xt+t^2-4x-6&=0 \\ 10x^2+(6t-4)x+t^2-6&=0 \\ x^2+\frac{(3t-2)x}{5}+\frac{t^2-6}{10}&=0 \\ \left(x-\frac{3t-2}{10}\right)^2-\left(\frac{3t-2}{10}\right)^2+\frac{t^2-6}{10}&=0 \\ x&=+\frac{3t-2}{10}\pm\sqrt{\left(\frac{3t-2}{10}\right)^2-\frac{t^2-6}{10}} && (*) \\ \end{align} \)

In order for the line \(3x+t\) to touch the circle in only one point, \(x\) has to be a singular solutions in \((*)\), i.e.

\( \begin{align} \left(\frac{3t-2}{10}\right)^2-\frac{t^2-6}{10}&=0 \\ \frac{9t^2-12t+4}{100}-\frac{t^2-6}{10}&=0 \\ 9t^2-12t+4-10t^2+60&=0 \\ -t^2-12t+64&=0 \\ t^2+12t-64&=0 \\ (t+6)^2-36-64&=0 \\ (t+6)^2-100&=0 \\ t&=-6\pm10 \\ y_1&=3x+4 \\ y_2&=3x-16 \\ \end{align} \)