This is from a math competition so it must not be something really long If a parabola touches the lines $y=x$ and $y=-x$ at $A(3,3) $ and $b(1,-1)$ respectively, then
(A) equation of axis of parabola is $2x+y=0$
(B)slope of tangent at vertex is $1/2$
(C) Focus is $(6/5,-3/5)$
(D) Directrix passes through $(1,-2)$
I thought the axis would be the angle bisector of the tangents passing through the focus but it turns out that is not the case in a parabola so how can I find anything..
On
(see figure below)
Parametric equations for the parabola are readily obtained, assuming a certain knowledge of quadratic Bezier curves : see for example the paragraph "Second order curve is a parabolic segment" in (https://en.wikipedia.org/wiki/B%C3%A9zier_curve)).
The tangents' intersection is clearly $O(0,0)$. Thus the parabola is nothing but the Bezier curve with endpoints $A$ and $B$ and "directing" point in $O$: $M=A(1-t)^2 + 2t(1-t)O + B t^2$, otherwise said:
$$\tag{1}\pmatrix{x\\y}=(1-t)^2\pmatrix{3\\3}+2t(1-t)\pmatrix{0\\0}+t^2\pmatrix{1\\-1}$$
$$\tag{2}\Leftrightarrow \ \ \ \ \cases{x=3(1-t)^2+t^2\\y=3(1-t)^2-t^2}$$
which constitutes a parametric description of the curve.
Eliminating $t$ between the two equations (2), one gets an implicit equation that can be written under the form :
$$\tag{3} \left(x-\frac{6}{5}\right)^2+\left(y+\frac{3}{5}\right)^2=\frac{(y+2x)^2}{5}$$
Taking square roots on both sides:
$$\tag{4} \sqrt{\left(x-\frac{6}{5}\right)^2+\left(y+\frac{3}{5}\right)^2}=\frac{|y+2x|}{\sqrt{5}}$$
(4) expresses the fact that the distance of $M(x,y)$ (the current point on the curve) to point $F(\frac{6}{5},-\frac{3}{5})$ is equal to the distance of $M$ to straight line $(D)$ with equation $y+2x=0$ (with slope $-2$). For distance formula see: (http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html). Thus $F$ is the focus of the parabola and $(D)$ its directrix.
Therefore, the parabola's axis, passing through $F$ and orthogonal to $(D)$ (thus with slope $\frac{1}{2}$) has the following equation: $y+\frac{3}{5}=\frac{1}{2}(x-\frac{6}{5})$, i.e., $y=\frac{1}{2}x-\frac{6}{5}.$
Conclusion:
propositions (A), (B) are false. For (B), it's not necessary to compute the tangent at the vertex because the slope of this tangent is the same as the slope of the directrix, which is $-2$, not $\frac{1}{2}$.
propositions (C), (D) are exact.
On
Conditions (A),(B),(C),(D) are not needed to determine the tilted parabola ( $xy$ term non-zero). Because out of five constants needed to determine a conic, one can be reduced as zero determinant for parabola.You have given two points and two slopes which are quite sufficient.
$$ (x, y, y^{ \prime} ) = (3,3,1), (1,-1,-1) $$
Parabola involving $xy,x^2, y^2,x,y $ terms can be solved for y in a quadratic as:
$$ y^2 - 2y ( ax+b) + (ax+b)^2 - (cx+d) =0 $$
$$ y = (ax +b) \pm \sqrt{ c x +d } $$
$\pm$ symbol separates regions on either side of vertical tangent. Plug in given point coordinates:
$$ 3 a + b+ \sqrt{ 3c+d} =3 , \quad a + b+ \sqrt{ c+d} = -1 ,\quad a + c \ /( 2 \sqrt{ c+d} ) = 1 ,\quad a - c \ /( 2 \sqrt{ c+d} ) = -1 $$
Using a CAS to reduce tedium we get $ (a,b,c,d) = ( 1/2, -3/4, 9/4,-27/16) $
The parabola is plotted to verify everything .It checks out (C), (D) but (A),(B) are inconsistent with first inputs and are clearly incorrect.
EDIT1
Some calculations needed (in progress)at focus and tangent orthogonal intersection on directrix
For parabola $ 4 a y = x^2 $
Points of tangency $$ (2at, a t^2)\quad (2a/t, a/t^2) $$ Point of intersection of polar chord tangents on directrix which happens at right angles at a point D $$ a [( t-1/t), -1] $$ Length of tangents $T_1,T_2$ given by $$ (T_1/a)^2 = t^4+ 3 t^2 + 1/t^2 +3 = 2, \quad (T_2/a)^2 = 1/t^4+ 3/ t^2 + t^2 +3 = 18 $$
$$\rightarrow t= \frac13 $$
$OF$ is perpendicular on hypotenuse AB
$$ \frac{1}{OF^2} =\frac{1}{2} + \frac{1}{ 9 \cdot 2} $$
$$ OF = \frac{3}{\sqrt 5} ... $$
$$ \cos \beta= \frac{OF}{OB} = \frac{ 3}{\sqrt10},\quad \sin \beta= = \frac{ 1}{\sqrt10}\quad $$
to be continued
On
It's well known that any rational quadratic Bézier curve is a conic section (ellipse, hyperbola, or parabola), and any (polynomial) quadratic Bézier curve is a parabola. There are fairly simple formulae for obtaining the geometric characteristics (directrix, focus, vertex) directly from the control points of the Bézier curve. For example, see the following paper and the earlier ones they cite:
Geometric Characteristics of Conics in Bézier Form
Cantóna, Fernández-Jambrina, Rosado María
Computer-Aided Design 43 (2011) 1413–1421
Here, we have a quadratic Bézier curve with control points $\mathbf{P}_0 =(3,3)$, $\mathbf{P}_1 = (0,0)$, $\mathbf{P}_2 = (1,-1)$. Several of the papers tell you that the axis of symmetry is in the direction of the vector $\tfrac12(\mathbf{P}_0 + \mathbf{P}_2) - \mathbf{P}_1$, which is $(2,1)$. Even without the papers, you can see this by letting $t\to\infty$ in the parametric equations of the curve. This shows that proposition (A) is false, and it follows immediately that (B) is also false.
Now that we know the axis direction, we can use the reflection property of a parabola to construct lines at $A$ and $B$ that pass through the focus. Intersecting these two lines gives us the focus. Or, use the formulae from the papers to get the focus.
I didn't write this up as a complete answer because I'm sure this is not how the problem was meant to be solved.
This figure illustrate the situation.
Using the fact that:
we can say that the focal chord has equation: $y=2x-3$, so we can test that the answer (C): $F=(6/5,-3/5)$ is correct. So we have the focus.
We can also use the fact that:
In the figure this mens that the two angles in $A$ and $D$ are equals and from this we can find the axis.
Finally, from the focus and the axis we can find the directrix as the line orthogonal to axis that passes thorough the common point of the two tangents : $C$.
If you o this you can verify that also the answer (D) is correct.
You can find the properties of the tangent used in this answer at: http://www.nabla.hr/CS-ParabolaAndLine2.htm.