Dumb Question: How does the fact that the circle x2+y2+2gx+2fy+c=0 passes through(a,0) leads to the condition -
a2+2ga+c=0-
Ans:- Since the circle passes through (a,0) the equation of circle must satisfy the point(a,0)-
so, a2+02+2g(a)+2f(0)+c=0-
0r a2+2ga+c=0 is obtained.-
let c1 and c2 be respectively the parabola x2=y-1 and y2=x p be any point on c1 and q be any point on c2. Let p1 and Q1 be reflections of P and Q-
respectively with respect to the line=y
prove that p1 lies on c2, Q1 lies on c1 and PQ> min[pp1,QQ1]-
Hence or otherwise determine points p0 and Q 0 on the parabolas c1 and c2 respectively. such that p0Q0 (Pq)-for all pairs of points (p,Q) with p on c1 and Q on c2.-
Ans:
Let co-ordinates of p and q are p(t, t2+1) and Q (s2 + 1, s) which lies on x2=y-1
and y2=x-1 respectively.-
p1 and Q1 be reflections of P and Q respectively with respect to the line y=x then-
we have, (PQ1)2=(t-s)2+(t2-s2)2
(p1Q)2
=>PQ1=PIQ
Thus PP1QQ1 is an isosceles trapezium
we have Pq> min {PP1,QQ1}
Let us take min {PP1,QQ1}=PP1
then (PQ)2=(pp1)2
=>(t 2+1-t)2+(t-t2-1)2=2(t2-t+1)2=f(t)say
we have f'(t)=4(t2-t+1)(2t-1)
NOw
Also
Hence f(t) is least when t=1/2 point p0 on c1 is and p1(which we take as Q0
and c2 are .Note that PQ for all pairs of(P,Q) with p on C1 and Q on c2.
Hence proved.
Dumb Question:- f' (t) is4(t2-t+1)(2t-1), but the only solution is t=1/2. what about the factor t2-t+1?
Ans:- Note that t2-t+1=
Now, this a positive quantity if t is + ve.
The coordinate plane is a real plane Where the points can take only real values and hence t has to real only. So, t2 - t + 1 cannot be zero.
=k (from the equation)
from (1) the equation of the axis of the parabola in x,y coordinate becomes-
The given parabola is x2=-8k(y-2k)........(9)
solving (8) and (9) we get
here D= 64k2-64t2(a2t2-k2
the axis given by(8) touches the given parabola.
Note:- if we take , the points of intersection of the axis and the given parabola will be imaginary.
Que:- A parabola drawn touching the axis of x at the origin and having its vertex at a given distance k from the x - axis . Prove that
the axis of parabola is a tangent to the parabola x2 + 8k(y - 2k) = 0.
Ans : Let the equation of the parabola be Y2 = 4ax
Any tangent to it at the point (at2, 2at) is
Yt = X +at2.........(1) The normal at the point( at2, 2at) is
Y + tX = 2at + at3........(2)
Take the equations of transformation -
in xy coordinates p=(0,0) and PT is the axis which is
tangent to the parabola at the origin.
Now,
the axis of the parabola(y=0) becomes-
the distance of the vertex v(0,0) in the x,y coordinates from pt -
p1 and Q1 be reflections of P and Q respectively with respect to the line y=x then-
and p1(which we take as Q0
.Note that PQ 
for all pairs of(P,Q) with p on C1 and Q on c2.
=k (from the equation)
, the points of intersection of the axis and the given parabola will be imaginary.
