Solve this problem that involves parametric equations

[chwala]

Homework Statement

##P## is a point on the parabola given by ##x=at^2, y=2at##, where ##a## is a constant. Let ##S## be the point ##(a,0)##, ##Q## be the point ##(-a,2at)##and ##T## be the point where the tangent at ##P## to the parabola crosses the axis of symmetry of the parabola.

(a) Show that ##SP=PQ=QT=ST=at^2+a##

(b) Prove that angle ##QPT## is equal to angle ##SPT##

(c) If ##PM##is parallel to the axis of the parabola, with ##M## to the right of ##P##and ##PN##is the normal to the parabola at ##P##, show that angle ##MPN## is equal to angle ##NPS##

Relevant Equations

Parametric equations

My take;
Part (a);

##\dfrac{dy}{dx}=\dfrac{1}{t}##

therefore,

##y-2at=\dfrac{1}{t}(x-at^2)##

##ty-2at^2=x-at^2##

##ty=x+at^2## implying that ##T## has co-ordinates ##(-at^2,0)##.

##SP=\sqrt{(a-at^2)^2+(0-2at)^2}##

##SP=\sqrt{4a^2t^2-2a^2t^2+a^2t^4+a^2}##

##SP=\sqrt{a^2t^4+2a^2t^2+a^2}##

##SP=\sqrt{a^2t^4+a^2t^2+a^2t^2+a^2}##

##SP=\sqrt{(at^2+a)^2}##

##SP=at^2+a##

also;

##PQ=\sqrt{(-a-at^2)^2+(2at-2at)^2}##

##PQ=\sqrt{(-a-at^2)^2}=\sqrt{a^2t^4+2a^2t^2+a^2}=at^2+a##

##QT=\sqrt{(-at^2+a)^2+(2at)^2}=at^2+a##

and

##ST=\sqrt{(-at^2-a)^2}=at^2+a##

thus shown.

For part (b);

Let the midpoint of ##QS## be denoted by ##K=\dfrac{-a+a}{2},\dfrac{0+2at}{2} =(0,at)##

##KP## is the perpendicular bisector to sides ##QP## and ##SP## respectively, we can use pythagoras theorem to show that
##PK^2+KS^2=PS^2## and ##PK^2+KQ^2=QP^2## this would imply that angle ##QPT=SPT##.

that is,

##PK=
\begin{pmatrix}
0 & \\
at & \\
\end{pmatrix}-
\begin{pmatrix}
at^2 & \\
2at & \\
\end{pmatrix}=
\begin{pmatrix}
-at^2 & \\
-at & \\
\end{pmatrix}
##

##\sqrt{PK^2+KS^2}##
##\sqrt{
\begin{pmatrix}
-at^2 & \\
-at & \\
\end{pmatrix}^2+
\begin{pmatrix}
a & \\
-at & \\
\end{pmatrix}^2}=PS
##

and

##\sqrt{PK^2+KQ^2}=##
##\sqrt{
\begin{pmatrix}
-at^2 & \\
-at& \\
\end{pmatrix}^2+
\begin{pmatrix}
-a & \\
at & \\
\end{pmatrix}^2}=QP
##
since ##PS=QP## and ##PK## is the perpendicular bisector to angle ##QPS## then it follows that angle ##QPT=SPT##.

 

[chwala]

...i should be able to use cos with vectors for part (b)...

Alternatively,

##QP=(at^2+a)i, KP=at^2i+atj, SP=(at^2-a)i+2atj##

Angle ##QPK=\dfrac{a^2t^4+a^2t^2}{at^2+a\sqrt{a^2t^4+a^2t^2}}=\dfrac{a^2t^2(t^2+1)}{(at)a(t^2+1)\sqrt{t^2+1}}=\dfrac{t}{\sqrt{t^2+1}}##

Angle ##SPK=\dfrac{a^2t^4-a^2t^2+2a^2t^2}{\sqrt{(at^2-a)(at^2-a)+4a^2t^2)⋅}\sqrt{a^2t^4+a^2t^2}}##

##=\dfrac{a^2t^4+a^2t^2}{at^2+a\sqrt{(a^2t^2(t^2+1)}}=\dfrac{a^2t^2(t^2+1)}{(at)a(t^2+1)\sqrt{t^2+1}}=\dfrac{t}{\sqrt{t^2+1}}##

Therefore angle ##QPK=SPK⇒QPT=SPT##

cheers!!

 

[SammyS]

...i should be able to use cos with vectors for part (b)...

Alternatively,

##QP=(at^2+a)i, KP=at^2i+atj, SP=(at^2-a)i+2atj##

Angle ##QPK=\dfrac{a^2t^4+a^2t^2}{at^2+a\sqrt{a^2t^4+a^2t^2}}=\dfrac{a^2t^2(t^2+1)}{(at)a(t^2+1)\sqrt{t^2+1}}=\dfrac{t}{\sqrt{t^2+1}}##
. . .

cheers!!

It looks like you are using the scalar product to find the cosines of the relevant angles. Initially when I read that first line, I thought you might be using the Law of Cosines.

In other words, what you are doing is: ##\displaystyle \ \ \cos(\angle QPK)= \dfrac{\overrightarrow{QP}\cdot\overrightarrow{KP}}{\left|{QP}\right|\,\left|{KP}\right|}##

You left the ##\cos## out of your expressions.
Also, in the first denominator, parentheses were missing from ##(at^2+a)## .

Now for some LaTeX:
Use \hat to get the "^" above i, j, k, etc. Also use \imath and \jmath to get rid of the dot above these letters.

Use \hat \imath to get ##\displaystyle \hat{\imath}##.

\angle gives the angle symbol. ##\displaystyle \angle QPK##