This is surely much better approach ankaj - But i think x and y should be such that they satisfy the circle's equation together. So, for y= 3+sqrt6 x=3
If we divide the circle into 4 parts(shift axis to 3,3) then we can see that the maximum value of y/x can be in quad 1 or quad 2(by observing we can eliminate quad 3 and 4, as y/x increases more rapidly in quad 1 as compared to quad 4, and in quad 3 y decreases while x increases) .
Just by little more observation we find that y/x will be maximum at y=3, x=3-sqrt6 (left most point on the circle). Good approach i must say :-)
just browsing thru this site today.... implex bhai......sahi qns hai... ne ways this is a nice qn n can be solved completely by graphical method... no need of using trigo/calculus... a simple qn actually.. draw a graph of the circle with the given eq. now we want to maximise y/x..... let y=kx now we choose a 'k' such that the graph of y=kx touches the circle. now the max such 'k' can be found if the said graph (y=kx) is tangent to the circle. it is tangent at 2pts one of which is max & the other minimum. if iv not made a calc mistake: the max y/x comes to (3+sqrt2)... nice qns..will visit this blog more frequently now ons.. bye
We can see that the mentioned equation represents a circle. Hence the maximum value of y/x is the greater of the slopes of the two common tangents from origin to the circle.
for pairs of real numbers 
which satisfy 
(x-3)^2 +(y-3)^2 = 6
ReplyDeletelet y = kx
then,
(x-3)^2 + (kx-3)^2 = 6
x^2+9-6x + k^2x^2 +9 -6kx = 6
(1+k^2)x^2 -(6+6k)x +12 = 0
Since x is real
D>=0
(6+6k)^2 - 48(1+k^2) >=0
-12-12k^2+72k>=0
-1-k^2+6k>=0
1+k^2-6k<=0
Max value of k will be between 5 and 6
5<k<6
Is this the correct approach?
no it is not good approach
ReplyDeletethe equation mentioned is of a circle with center at 3,3 and radius sqrt(6)
ReplyDeleteso y max= 3+sqrt(6) x min= 3-sqrt(6)
y/x= 5+2sqrt(6)
is this approach, fine???
This is surely much better approach ankaj - But i think x and y should be such that they satisfy the circle's equation together. So, for y= 3+sqrt6 x=3
ReplyDeleteIf we divide the circle into 4 parts(shift axis to 3,3) then we can see that the maximum value of y/x can be in quad 1 or quad 2(by observing we can eliminate quad 3 and 4, as y/x increases more rapidly in quad 1 as compared to quad 4, and in quad 3 y decreases while x increases) .
Just by little more observation we find that y/x will be maximum at y=3, x=3-sqrt6 (left most point on the circle). Good approach i must say :-)
Hence, y/x= 3 + sqrt6.
we can think of it sin x ->1
ReplyDeletecos x->0
so
3+root6 /3 = approx2.. something less than 2
y/x =3+root6 sint/3+root6cost
ReplyDeletedifferenciatin i got
sint +cost = 2/root6
y/x = 7+root5 /7-root5
just browsing thru this site today.... implex bhai......sahi qns hai...
ReplyDeletene ways this is a nice qn n can be solved completely by graphical method... no need of using trigo/calculus... a simple qn actually..
draw a graph of the circle with the given eq.
now we want to maximise y/x..... let y=kx
now we choose a 'k' such that the graph of y=kx touches the circle.
now the max such 'k' can be found if the said graph (y=kx) is tangent to the circle. it is tangent at 2pts one of which is max & the other minimum.
if iv not made a calc mistake: the max y/x comes to (3+sqrt2)...
nice qns..will visit this blog more frequently now ons..
bye
yupp but using coordinate, we alwasy know what are coods of points on a circle, we are done without any knew work
ReplyDeleteWe can see that the mentioned equation represents a circle. Hence the maximum value of y/x is the greater of the slopes of the two common tangents from origin to the circle.
ReplyDeleteHence max(y/x) = tan(45+arcsin(sqrt(6)/3sqrt(2))
tan(arcsin(1/sqrt(3)) = 1/sqrt(2)
Therefore, max(y/x) = 3+2sqrt(2)
vijay is the best........!!!
ReplyDelete