Illustration 10.
The general equation to a system of parrallel chords in parabola  is 4x - y + K = 0 . What is equation of corresponding diamcter ?
Ans Equation of parabola is y
And the equation of system of | | chords
is 4x = y + K = y .
 m = 4 .
So, diameetcr 
or 56y = 25.
Some Important and useful results
1. Focal distance of a point (at2,2at) is a + at2
Why?
Now focal distance = Pf = PB = a + at2
= a +x,sub>1 (if (s1, y1) is point )
2. Foot of perperdicilar drawn from focus of parabola to any tangent will lie on tangent at vertex.
Why?
Equation of tangent is y = mx + a/m ---(1)
Equation of normal is 
or my + x = a ----(2)
Solving (1) & (2) we get
x = 0, y =a/m .
(O,a/m) is the point and clearly it lies or y - axis
3. The circle descvibed on any focal chord as diamcter touches the directrix of the parabola .
OR.
Any two tangents drawn from point on the directrix to parabola are perpendicular to each other and the chord of contact is focal chord .
4.) If any focal chord meets parabola at t1and t2.
then t1t2 = - 1 .
Why ?
Slope of AC = slope of AB
So,
=>t1(1 - t22) = t2(1 - t12)
t1- t1t2(t2) = t2 - t1t2 (t1)
(t1 -t2) + t1t2(t1- t2) = 0
(t1 -t2)( 1 + t1t2) = 0
=> t1t2 = - 1.
Dumb Question Why t1 - t2  0?
Ans If t1 -t2 = 0 => t1 = t2
So, this will mcan that point B and C are same point which is not true .
Hence t1 - t2 0
5). If any chord joining t1 and t2 subtends right angle at vertex that t1t2 = - 4 .
Why?
Slope of VA  slope of VB = - 1 .
t1t2 = - 4 .
6). Normal at t meets parabola at point (- t -2/t)
Why ?
Normal at t is ly = - tx + 2at + at3
It meets parabola at t1
t1 will also satisfy the equation
2ar1 = - t (at'2) + 2at + at3
or t(t'2 - t2) + 2 ( t'-t) = 0
or (t'-t) (t(t' + t) + 2) = 0
=> t' = t or t' = - t -2/t
But t' t as lpoint is different . So, t' = - t-2/t
7) If two normals at point t1 and t2 meet again on parabola then t1t2 = 2 .
Why ?
Now  (from previous result) and also
=> t1t2 = 2
8) Point of intersection of tangents at t1 and t2 is (at1 t2 ,a (t1= t2)).
9) Point of intersection of normal drawn at t1 and t2 is
[2a + a (t12+t22 + t1 t2), - at1t2(t1 + t2)].
(10) Reflection property of parabola. the tangent (pT) and normal (PN) ofparabola y2=4ax at p are the internal and external bisectors of SPM and Bp is  to axis of parabola and BPN=SPN diagram 25
Illustration:
A ray of light coming along line y=b from the positive direction of x-axis strikes a concave mirror whose intersection with x-y plane is a parabola y2= 4ax. Find equation of reflected ray and show it passes through focus of parabola
. both a and b are positive.
Ans:- given parabola is y2=4ax.
Equation of tangent at
diagram26 
slope of tangent=
hence, slope of normal 
slope of reflected ray
=tan(180-2 )
=-tan 2
=(y-b)(4a2-b2)=-(4ax-b2)b
which clearly passes through focus s(a,0)
Easy :-
E_1 show that the line xcos +ysin=p
touches the parabola y2
=4ax if p cos+a sin2=0
and that the point of contact is (atan2 ,-2atan)
Solution: The given line is
xcos +ysin=p
or y=-xcot+pcosec
 m=-cot and c=p cosec
since the given line touches the parabola
c=
or cm=a
or(pcosec)(-cot=a
and point of contact is
E-2 show that normal to the parabola y2=8x at the point(2,t) meets
it again t (18,-12). find also the length of the normal chord.
solution-
s comparing the given parabola(i.e, y2=8x) with =4ax
4a=8
a=2
since normal at (x1,y1) to the
parabola y2=4ax is
y-y1=
Here x1=2 and y1=4
equation of normal is,
y-4=
= y-4=-x+z
=y+x-6=0............(1)
diagram solving (1) and y2 =8x
y2=8(6-y)
=y2+8y-48=0
(y+12)(y-4)=0
y=-12 and x=2
hence point of intersection of normal and parabola are(18,-12) and(2,4) therefore
normal meets the parabola at(18,-12)
and length of normal chord is distance between their points
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