Question 1157992
(a) Sketch the parabola, and make calculations that confirm that P = (7, 12) is on it.
{{{drawing(300,300,-4,16,-6,14,
graph(300,300,-4,16,-6,14,sqrt(8x+8)+4,4-sqrt(8x+8)),
blue(triangle(-3,-6,-3,14,-3,4)),circle(7,12,0.2),
locate(2,-4.5,blue(directrix)),locate(7.1,12,P(7,12)),
blue(arrow(1.9,-5,-3,-5)),circle(1,4,0.2),
locate(1.1,4,F(1,4)),locate(5,5,parabola),
red(arrow(5.6,5.1,5.6,11.27)),
green(arrow(5.6,4.1,5.6,-3.27))
)}}}
A parabola is the locus of the points that are at the same distance from the directrix and the focus.
The distance between directrix {{{x=-3}}} and {{{P(7,12)}}} is
{{{7-(-3)=7+3=10}}}
The distance between focus {{{F(1,4)}}} and {{{P(7,12)}}} is
{{{sqrt((7-1)^2+(12-4)^2)=sqrt(6^2+8^2)=sqrt(36+64)=sqrt(100)=10}}}
{{{P(7,12)}}} is at the same distance from the directrix and the focus,
so it is on the parabola.
We could prove P is on the parabola from the equation of the parabola.
Knowing that the equation of a parabola with its vertex at the origin and focal distance {{{f}}} is
{{{x=(1/4f)y^2}}} for parabolas with the x-axis as an axis of symmetry,
we can translate that equation for a parabola with the vertex at (-1,4), halfway between focus and directrix.
Doing that, we found that our parabola has {{{f=1-(-1)=2}}} and the equation for our parabola would be  
{{{x-(-1)=(1/(4*2))(y-4)^2}}}-->{{{x+1=(1/8)(y-4)^2}}}
For {{{y=12}}} we get {{{x+1=(1/8)(12-4)^2=(1/8)*8^2=8}}}-->{{{x=8-1=7}}}
  
(b) Find the slope of the line μ through P that is tangent to the parabola.
We can estimate the slope of the tangent line from the graph or calculate it from the derivative of the function.
From {{{x+1=(1/8)(y-4)^2}}}-->{{{x=(1/8)(y-4)^2-1}}} we can find the derivative
{{{dx/dy=(1/8)*2(y-4)}}} .
For {{{P(7,12)}}} , that derivative is {{{dx/dy=(1/8)*2(12-4)=(1/8)*2*8=2}}}
For the function graphed in red above, {{{dy/dx=highlight(1/2)}}} is the derivative and slope of the tangent at point P.
 
(c) Calculate the size of the angle that μ makes with the line y = 12.
(d) Calculate the size of the angle that μ makes with segment F P . 
{{{drawing(300,300,-4,16,-2,18,
graph(300,300,-4,16,-2,18,sqrt(8x+8)+4,4-sqrt(8x+8),x/2+8.5),
circle(7,12,0.2),locate(-3.9,6.5, blue(mu)),
locate(7.1,12,P(7,12)),triangle(1,4,7,12,4,8),
circle(1,4,0.2),triangle(-4,12,7,12,2,12),triangle(-12,12,7,12,16,12),
locate(1.1,4,F(1,4)),locate(-3.9,12,y=12),
green(arc(7,12,16,16,-26.6,0)),locate(13,14,green(alpha)),
green(arc(7,12,13,13,153.4,180)),locate(1,11,green(alpha)),
green(arc(7,12,7,7,126.9,180)),locate(4.5,11,green(beta))
)}}}
The line {{{y=12}}} slope is zero, it is parallel to the x-axis.
Line {{{blue(mu)}}} with slope {{{1/2}}} makes an angle {{{green(alpha)}}} with the x-axis and with the line{{{y=12}}} such that
{{{tan(green(alpha))=1/2}}} --> {{{green(alpha)=highlight(26.565^o)}}}
Segment FP, connecting {{{F(1,4)}}}} and {{{P(7,12)}}} , and line FP, have a slope of
{{{(12-4)/(7-1)=8/6=4/3}}} . A line, or segment with such a slope would make an angle {{{green(beta)}}} with the x-axis and line {{{y=12}}} such that
{{{tan(green(beta))=4/3}}} --> {{{green(beta)=53.130^o}}}
The angle that μ makes with segment FP is
{{{green(beta)-green(alpha)=53.130^o-26.565^o=highlight(26.565^o)}}}