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. $\displaystyle \angle QPK$