We give F coordinates (p/2,0).
P(x,y) is on the parabola <=> |P,d| = |P,F| <=> |P,d|2 = |P,F|2 <=> p 2 p 2 (x + -) = (x - -) + y2 2 2 <=> ... <=> y2 = 2p xThe point F is called the focus and the line d is the directrix. The axis of symmetry of the parabola is called the axis of the parabola. The axis of the parabola is the line through the focus and perpendicular to the directrix. The intersection point of that axis and the parabola is called the vertex of the parabola.
Then we have a parabola with an equation
x2 = 2 p y <=> 1 y = ----- x2 2 p Now the focus is (0, p/2) and the directrix is y = -p/2. Let a = 1/ (2 p) Now the equation is y = a x2The focus is ( 0, 1/(4a) ) and the directrix is y = - 1/(4a)
Since the equation is y = ax2 , we can find the slope of the tangent line by means of the derivative. In point P(xo, a xo2) of the parabola the slope of the tangent line is 2 a xo.
However, the parabola was defined as a set of points which satisfy a geometric condition. Now, we set up the equation of a parabola with focus F(a,b) and directrix ux+vy+w=0.
Point P(x,y) is on the parabola if and only if |P,d| = |P,F|.
P(x,y) is on the parabola <=> |P,F| = |P,d| <=> |P,F|2 = |P,d|2 <=> (x - a)2 + (y - b)2 = (l x + m y + n)2 (1)Example:
We calculate the equation of the parabola with focus F(1,2) and directrix d: 3x+4y-2=0.
The directrix has normal equation (3/5) x + (4/5)y - 2/5 = 0.
The equation of the parabola is
(x-1)2 + (y-2)2 = ((3/5) x + (4/5)y - 2/5)2 (2)It is obvious that this form can be simplified to
16 x2 - 24 xy + 9 y2 - 38 x - 84 y + 121 = 0 (3)The annoying thing is that that form not shows where the focus is and where te directrix is. If we want F and d, we have to calculate F en d from the last expression. Therefore we bring (1) in another shape.
(x - a)2 + (y - b)2 = (l x + m y + n)2 <=> x2 - 2ax + a2 + y2 -2by + b2 = l2 x2 + m2 y2 + n2 + 2 l.m xy + 2 l.n x + 2m.n y <=> (1-l2)x2 - 2 lm xy + (1-m2) y2 = 2x(a + l.n) + 2y(b + m.n) -a2 - b2 + n2 <=> (m x - l y)2 = 2x(a + l.n) + 2y(b + m.n) -a2 - b2 + n2 (4)We transform (3) to the shape (2).
16 x2 - 24 xy + 9 y2 - 38 x - 84 y + 121 = 0 <=> (4 x - 3 y)2 = 38 x + 84 y - 121 Since l2 + m2 must be equal to 1, we divide both sides by 25 <=> ( (4/5) x - (3/5) y)2 = 38/25 x + 84/25 y - 121/25 We compare this result with (4) and we see that m = (4/5) en l = (3/5) a + l.n = 19/25 (5) b + m.n = 42/25 (6) n2 - a2 - b2 = - 121/25 (7) Since l and m is known, we bring a and b from (5) and (6) to (7) and then a = 1 ; b = 2 ; n = -2/5. The equation (3) is a parabola with focus F(1,2) and directrix (3/5)x + (4/5)y -2/5 = 0 <=> 3x+4y-2=0.
We start with a fixed coordinate system, an arbitrary parabola P and a fixed parabola P" with equation y = x2.
There is always a suitable rotation and translation such that P is transformed in
a parabola P' with equation y = ax2 with a > 0.
So, P is similar to P'. We also know that a homothetic transformation transforms a figure in a similar figure. We choose O(0,0) as center of a homothety h and we choose 'a' as factor.
The transformation formulas are
h (x,y) ----> (ax, ay) Point D' is on parabola P' <=> D'(x , a x2) h transforms this point D' in the point D" D"( a (x) , a (a x2)) = D"( (a x) , (a x)2 ) <=> Point D" is on the parabola P" with equation y = x2 So, the arbitrary parabola P is similar to the fixed parabola P".Thus all parabolas are similar to the same fixed parabola.
A parabola is completely determined by a given focus and directrix.
Consider two parabolas :
P1 with focus F1 and directrix d1 and P2 with focus F2 and directrix d2 The parabolas P1 and P2 are congruent <=> There is a displacement v such that v(F1) = F2 and v(d1) = d2 <=> The relative position of F1 against d1 is the same as The relative position of F2 against d2 <=> The distance from F1 to d1 = the distance from F2 to d2 <=> The distance from F1 to the tangent at the vertex of P1 is equal to The distance from F2 to the tangent at the vertex of P2Example:
The parabola P1 with equation y = 0.25 x2 is congruent with the
parabola P2 with directrix d2 : x - y = 0.|
The focus of P2 is on the line b with equation x + y - 1 = 0.
Find all possible equations of P2.
The parabola P2 is congruent with P1 if and only if the focus F2 lies at a distance 2 from the directrix x - y = 0.
Let F2 be a variable point of the line b. We use parameter t.
F2 = F2(t , 1-t).
| F2, d2| = 2 <=> | 2 t -1| ------------- = 2 sqrt(2) <=> (2t - 1)2 = 8 <=> t = 0.5 ± sqrt(2)So there are two possible locations for the F2.
F2 = F2(0.5 + sqrt(2) , 0.5 -sqrt(2) ) The equation of the parabola P2 is (x - 0.5 -sqrt(2))2 + (y - 0.5 + sqrt(2))2 = (x-y)2/2 <=> x2 + 2 x y + y2 - 7.66 x + 3.66 y + 9 = 0
Solve this case as exercise
x = 2 p t2 (1) y = 2 p t (2)The real number t is the parameter.
y 2 x = 2 p (---) <=> y2 = 2 p x 2 pHence, the two associated lines constitute a curve and that curve is the parabola.
D( 2 p t2 , 2 p t)is on the parabola for each t-value and with each point of the parabola corresponds a t-value.
y2 = 2p xTo obtain the slope of the tangent line we differentiate implicitly.
2 y y' = 2 p <=> y' = p/ySay D(xo,yo) is a fixed point of the parabola.
p --- y0The equation of the tangent line is
p y - y0 = -- (x - x0) y0 <=> y0 y - y02 = p x - p x0 Since y02 = 2p x0 <=> y0 y - 2 p x0 = p x - p x0 <=> y y0 = p (x + x0)The last equation is the tangent line in point D(x0,y0) of a parabola.
y2 = 2 p x y = m x + q Substitution gives (m x + q)2 = 2 p x <=> m2 x2 + 2 (m q - p) x + q2 = 0 The line t is a tangent line if and only if the roots of the last equation are equal. Therefore the discriminant has to be zero. 4 (m q - p)2 - 4 m2 q = 0 <=> 4 p (p - 2 m q) = 0 <=> p q = --- 2 m The tangent line with a given slope m is p y = m x + --- 2 m
y - y0 = m(x - x0)The intersection points with the parabola are the solutions of the system
y2 = 2 p x y - y0 = m(x - x0) We substitute x from the first equation into the second one. y2 y - y0 = m(--- - x0) 2 p <=> - m y2 + 2 p y - 2 p y0 + 2 p m x0 = 0The line t is a tangent line if and only if the roots of the last equation (in y) are equal. Therefore the discriminant has to be zero.
4 p2 + 4 m (2 p m x0 - 2 p y0) = 0 <=> 2 x0 m2 - 2 y0 m + p = 0The roots of this equation are the slopes of the two tangent lines.
y0 + sqrt(y02 - 2 p x0) m1 = ------------------------------ 2 x0 y0 - sqrt(y02 - 2 p x0) m2 = ---------------------------- 2 x0The equations of the tangent lines are
y2 y - y0= m1(----- - x0) ; p y2 y - y0= m2(----- - x0) pThe two lines are orthogonal if and only if m1 . m2 = -1
p <=> ---- = -1 2 x0 -p <=> x0 = ---- 2From this we see that if point P is on the directrix, the tangent lines are orthogonal.
|The directrix is the locus of the points P such that the tangent lines through P to the parabola are perpendicular.|
The given solution is not 'the' solution.
Most exercises can be solved in different ways.
It is strongly recommended that you at least try to solve the problem before you read the solution.
Find the points of the parabola y2 = 4x such that the distance from these points to the focus is four.
The parabola P has equation y2 = 2 p x.|
A variable point P has coordinates (p/2,t). The parameter is the real number t greater than p .
Calculate the the tangent of the sharp angle between the tangent lines through P.
The tangent lines t and t' in points P(x1,y1) and in P'(x2,y2) of a parabola
Prove that y1.y2 = - p.p
|Point P is on a parabola. The projection of P on the axis of the parabola is Q. The normal line in P intersects the axis in point N. Show that |QN| is constant.|
|The line y = x+3 is a tangent to the parabola y2 = 2px. Calculate p.|
|The line y = 0.5 x - 4 is a normal line to the parabola y2 = 2px. Find the tangent line to the parabola corresponding with that normal line.|
|We consider two points P and P' on a parabola. The tangent lines in these points intersect on the directrix. Find the product of the ordinates of P and P'.|
|Consider the tangent lines in 4 points of a parabola. They intersect and form a quadrilateral. Show that the line through the midpoints of the diagonals of the quadrilateral is parallel to the axis of the parabola.|
|Take four points on a parabola. Find the condition in order that the four point are concyclic.|
|We know that x2 + y2 = (x - c)2 is the equation of a set confocal parabolas with focus F(0,0). Q(r,s) is a point different from F. Show that there are exactly two parabolas of the set through Q. Show that there is an orthogonal intersection of these two parabolas in point Q.|
|Consider a variable tangent line to the parabola y2 = 2 p x. This line cuts the x-axis at point Q and the y-axis at point R. Find the locus of the center M of [QR].|
A variable line l has a fixed slope m. The line l intersects the parabola y2 = 2px
in the points Q and R.|
Show that the center M of [QR] lies on a fixed line parallel to the axis of the parabola.
|P is the parabola y2 = 2p x. We connect a variable point P of the parabola with the vertex of the parabola. Show that the locus of the center M of this variable chord PO is a parabola.|
|A parabola has an equation x2 + y2 = (x+1)2. Find the focus F and the directrix. A variable line through F intersects the parabola in the points A and B. Find the locus of the center M of [AB].|
A is a variable point of the parabola y2=6x and B(2,8) is a fixed point.|
Find A such that |AB| is minimum.
|The points A and B are on the parabola y2 = 2p x. M is the center of the segment [A,B]. The lines r an r' are the tangent lines in A and B to the parabola. The line s is parallel to the axis of the parabola and contains M. Show that the lines r, r' and s are concurrent lines.|
Find t such that the following parabolas are congruent.|
P1 : y2 = t x with t > 0
P2 : (x - 1)2 + (y - 2)2 = (0.6 x + 0.8 y)2