# Conics, tricks and techniques to avoid polar theory and differential calculus

In the present article we want to give formules and solve problems, that are generally teached after the introduction of derivation and successively polar theory, avoiding them, In this way many such problems are reachable to students of lower school level.

We recall a known trick to obtain metric and focal properties of conics that are usually demonstrated after the introduction of polars and their property. These are usually teached during university courses while the present method can be teached in secondary schools. The trick consists in a simple by absurd reduction argument.

For example we demonstrate the following:

Theorem: The tangent line to a parabola at a point P on it bisects the joining PF of the point to the focus and the perpendicular PH from P to the directrix.Moreover it passes through the middle point M of FH that coincides with the intersection of FH and the tangent to the vertex.

Dim: Given the parabola $\Gamma$, F its focus and d its directrix, let P be a point on $\Gamma$. Let’s draw PH perpendicular to d, where H is its foot. Let’s draw the bisector of $\widehat{FPH}$. If by absurd it wouldn’t be tangent to the parabola it would meet it in a point Q different from P. From Q let’s draw the perpendicular QK to d where K is its foot. Consider triangle(FPQ) and triangle(PQH): PQ is in common, $\widehat{FPQ}=\widehat{QPH}$ by hypothesis, FP=PH by the characteristic property (here definitory) of the parabola $\Rightarrow$triangle(FPQ)=triangle(PQH)$\Rightarrow$ FQ=QH. But FQ=QK by the definitory property of the parabola at Q $\Rightarrow$ QK=QH, that is a cateto = hypothenuse. This is possible only if Q is on the vertical PH that is clearly impossible. Then the bisector of $\widehat{FPH}$ cannot meet $\Gamma$ elsewhere from P and hence is tangent to it.
The triangle FPH is isosceles and so its bisector, the tangent, is also median and so passes through M middle point of FH. But this coincides with the intersection of FH eith the tangent at the vertex, because this latter is parallel to the directrix and V is equidistant from F and the diectrix.

And with the same arguments we can demonstrate all metric and focal properties of conics.

Property: the parabola is convex at its vertex.

proof: given a parabola $\Gamma$ and its vertex V, the tangent t to the vertex and the directrix d we proof that no point P of $\Gamma$ can lie between t and d or on the hyperplane generated by d not containing V. By absurd let a point P of $\Gamma$ lying between t and d, not on FH. Let F be the focus and FV be the perpendicular to  t meeting d in H. From P let draw the perpendicular to FH. By hypotheses k lies between V and H. Then we have $PF\gneq FK$ because FP is hypotenuse of FKP, $FK\gneq KH$ because k is between V and H and FV=VH. Then we have $PF\gneq KH$ where the later is the distance from P to the directrix. But this cobtradicts the defining property of the points of $\Gamma$. Similarly for points on FH or on the hyperplane generated by d not containing V

Equation of the tangent to the parabola in a point P on it:
From theorem 1) we have that such tangent passes through the points $P({ x }_{ 0 },{ y }_{ 0 }=a{ x }_{ 0 }^{ 2 }+b{ x }_{ 0 }+c)$ and $M(\frac { { x }_{ 0 }-\frac { b }{ 2a } }{ 2 } ,\frac { -{ b }^{ 2 }+4ac }{ 4a } )$ (its abscissa is midway from that of V and P and lies on the tangent at V). Easily we obtain:
$y-(a{ x }_{ 0 }^{ 2 }+b{ x }_{ 0 }+c)=(2a{ x }_{ 0 }+b)(x-{ x }_{ 0 })$  and above all  $m=2a{ x }_{ 0 }+b$

Equations of tangents drawn to a parabola from a generic point (easy to memorize):
Given the parabola $y=a{ x }^{ 2 }+bx+c$ and the point $P({ x }_{ 0 },{ y }_{ 0 })$ we write the generic line through P $y-{ y }_{ 0 }=m(x-{ x }_{ 0 })$, we solve  the system between this and the parabola and we arrive to the Resolvent Equation $a{ x }^{ 2 }+(b-m)x+(c+m{ x }_{ 0 }-{ y }_{ 0 })=0$. Imposing the condition of tangence $\delta = 0$ we find ${(b-m)}^{2}-4a(c+m{x}_{0}-{y}_{0})=0$. Solving this equation we have the two values for m:
${ m }_{ 1,2 }=b+2a{ x }_{ 0 }\pm 2\sqrt { a(a{ x }_{ 0 }^{ 2 }+b{ x }_{ 0 }+c-{ y }_{ 0 }) }$
This formula is easy to memorize: the first term is simple and under the root we observe that the long factor is the equation of the parabola reversed in sign and calculated in the point P!
Observe that:
${ m }_{ 1,2 }$ are real $\Leftrightarrow a(a{ x }_{ 0 }^{ 2 }+b{ x }_{ 0 }+c-{ y }_{ 0 })\ge 0 \Leftrightarrow \begin{cases} a0\quad \wedge \quad P\quad under\quad the\quad parabola \end{cases}$
$m=2a{ x }_{ 0 }+b \Leftrightarrow a(a{ x }_{ 0 }^{ 2 }+b{ x }_{ 0 }+c-{ y }_{ 0 })=0 \Leftrightarrow$ P lies on the parabola
2nd proof: we reverse the order of ideas. We know that m is tied with $P({ x }_{ 0 },{ y }_{ 0 })$ by an equation of the form $A({ x }_{ 0 },{ y }_{ 0 }){m}^{2}+B({ x }_{ 0 },{ y }_{ 0 })m+C({ x }_{ 0 },{ y }_{ 0 })=0$ where A,B,C are polinomials in the indeterminates ${ x }_{ 0 } and { y }_{ 0 }$. In fact all the systems and conditions give  rise only to algebraic relations between the variables $m, { x }_{ 0 }, { y }_{ 0 }$.The relation is of second degree in m because, given a point $P({ x }_{ 0 },{ y }_{ 0 })$ there are two tangents through it and so two values of m (if real). But the coefficients must all be of first degree. Infact given a value m0 for m there is only exactly one tangent to the parabola with that angular coefficient: its points $P({ x }_{ 0 },{ y }_{ 0 })$ are exactly the points of the plane that give that value of m0. This means that subsituting m0 for m in the equation it gives the equation of that tangent and so A,B,C must be linear.
The coefficient A is a constant because we have always two tangents (if real) and never one only (there is no vertical tangent). If A were of the form ${a}_{0}{ x }_{ 0 }+{a}_{1}{ y }_{ 0 }+{ a }_{ 2 }$ then the points that make it equal to 0 would have only one m. We can divide by this constant and suppose without loss of generality that A=1.
The equation C=0 gives the points of the plane having one of the two tangents in orizzontal position. But this locus is the tangent at the vertex $y+\frac { \Delta }{ 4a } = 0$ and then we have $C=k(y+\frac { \Delta }{ 4a } )$, k constant.
Then we are reduced to ${m}^{2}+({b}_{0}{ x }_{ 0 }+{b}_{1}{ y }_{ 0 }+{ b }_{ 2 })m+k({y}_{0}+\frac { \Delta }{ 4a } )=0$.
Now the discriminant of this wquation is the locus of points with two coincident tangents and is then the parabola itself:
${({b}_{0}{ x }_{ 0 }+{b}_{1}{ y }_{ 0 }+{ b }_{ 2 })}^{2}-4k({y}_{0}+\frac { \Delta }{ 4a } )=t(a{ x }_{ 0 }^{ 2 }+b{ x }_{ 0 }+c-{ y }_{ 0 })$
cpmparing coefficients we obtain: ${b}_{1}=0, k=\cfrac{{b}_{0}^{2}}{4a},{b}_{2}=\cfrac{{b}_{0}b}{2a}$ and substituting we arrive at:
${m}^{2}+({b}_{0}{ x }_{ 0 }+\cfrac{{b}_{0}b}{2a})m+\cfrac{{b}_{0}^{2}}{4a}({y}_{0}+\frac { \Delta }{ 4a } )=0$
changing ${b}_{0}=2a{b}_{3}$ we get:
${m}^{2}+{b}_{3}(2a{ x }_{ 0 }+b)m+a{b}_{3}^{2}({y}_{0}+\frac { \Delta }{ 4a } )=0$
When ${y}_{0}=-\frac { \Delta }{ 4a }$ the point $M({ x }_{ 0 },{ y }_{ 0 })$ is on the tangent at the vertex. We know from the demonstration of the tangent equation with P on the parabola that the corresponding point of tangence (different from the vertex) had ${x}_{P}=2{x}_{0}+\cfrac {b}{2a}$ and the corrisponding angular coefficient was $2a{x}_{P}+b=2a(2{x}_{0}+\cfrac {b}{2a})+b=4a{x}_{0}+2b$.
Subsituting ${y}_{0}=-\frac { \Delta }{ 4a }$ in the equation of m we must find the solutions m= 0 and $m=4a{x}_{0}+2b \Rightarrow {b}_{3}=-2$ and we arrive at:
${m}^{2}-2(2a{ x }_{ 0 }+b)m+(4a{y}_{0}+ \Delta )=0 \Rightarrow { m }_{ 1,2 }=b+2a{ x }_{ 0 }\pm 2\sqrt { a(a{ x }_{ 0 }^{ 2 }+b{ x }_{ 0 }+c-{ y }_{ 0 }) }$ as before.
References

Federigo Enriques: Lezioni di Geometria proiettiva, Zanichelli
E. Marchionna: Appunti ed esercizi di Geometria II, Masson