1Thank You all for the comments
Asish has shown a very nice method to function Q....
3.>A circle has diameter being any focal chord of the conic y2 -4x-y-4=0. Find the equation of line it will always touch
.
PHysics:
1>A motorcycle is going on a overbridge of radius R.The bike is acsending the overbridge with constant speed,then the normal rxn on it :
a)increases b)decreses c)remains same D) fluctuates
2>A rod of length L is pivoted at 1 end and is rotated with uniform ω .let T1 an dT2 be tensions at pts L/4 and 3L/4≤.Then THe inequality between T1 and T2 is:??
Old Question:
3> Train A runs frm east to west,pressing the track with force F1 and Train B,pressing the tack with F2(of same mass) frm west to east at same speed along equator:
a)F1>F2 b)F!<F2 c)F1=F2
PLZ GIVE THE LOGIC ALSO.
CHEM:How to know that a rxn proceeds in forward direction by looking at it??
for e.g:
a)SnCl4+Hg2Cl2--->SnCl2+2 HgCl2
b)NH4Cl+ NaOH-----> water+ ammonia +NaCl
Maths:
The population rises @ 5% every yr. In how many yrs will it double itself? [how to solve it using differntial eqn]
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38 Answers
1
NOW I CAN't explain more tht porblem is ILL FRAMED....if u cn see tht good for u else i m LEAST BOTHERED NOW....!
1
39y² - y = 4x + 4
=> y² - y + 1/4 = 4x + 17/4
=> (y - 1/2)² = 4(x + 17/16)
A parabola with X-axis symmetry, vertex at (1/2, -17/16).
Let any two points on the parabola be P(at1, 2at12) and Q(at2, 2at22).
Any chord formed by such a pair of points would be of the form :
y(t1 + t2) - 2x = 2at1t2
Now a = 1, t1t2 = -1(focal chord)
=> y(t1 + t2) - 2x = -2
This is also the diameter of the circle.
Not really sure what to do after this...
1even i am unable to reach the answer are u sure y(t1 + t2) - 2x = -2
is the diameter of the circle ???
..neway thanks for trying
4>If the line y-√3+3=0 cuts the parabola y2=x+2 at A and B ,then PA.PA is equal to: ??
5>If the chord of conatct of tangents from a pt. P to the parabola y2=4ax touches parabola x2=4by ,then the locus of P is:
a)circle, b)parabola c)ellipse d)hyperbola
14) question is wrong LOL!!!!!!!! [question post karne se pehle aur uspe argue karne se pehle ek baar socha karo...a line of form y= a cn ONLY INTERSECT ONCE a parabola of form y2=4cx...]
bhaiyyeeeeee P kahan se tapka in the problem yeh P hai kaun???
[EXCLUSIVELY FOR MACHINE TYPE PROBLEM SOLVERS:::
NOTE:::kuch HOSHIYAARON KO y ke TWO VALUES milenge LOL!!!!![by taking square root.but one of those values won't satisfy the GIVEN LINE]
39Arey babua method bhi dikhaiye...hum aapke tarah BRILLIANT nahi hain :P
But if I have any common sense, that line in Q3 is y - √3x + 3 = 0 and we have to find PA.PB
1Q3 is solved!!!
The concept here is :Yhe circle has has ffocal chord as diameter n touches dirctrix..
parabola:=> (y - 1/2)² = 4(x + 17/16)
so
(x + 17/16)=-1
or x=-33/16..ans
Anand is right!!!
PLz try Q 4 and Q5..
@anand: all qs have been typed with utmost care, so no mistakes
1@pririths exctly thts wat i m saying to avinav since long but he says the QUESTION MUST BE SOLVED EXACTLY IN FORM HE GAVE ....no modifications allowed......so u see at the most COMMON SENSE IN SUCH CASE CAN TELL THT QUESTION IS WRONG!!!...so don't expect me to use my senses to change it m afraid someday avinav will make a line of form y=a to intersect a parabola of form y2=4ax twice :)
1yaar avinav plz do reply in time so tht if we cn know its right or wrong n try to make paint it neat files n upload solution...u know upladoing it takes time in MY PATHETIC SERVER TOO[pritish plz no elaboration fo my net [3][3][3][3][3]]
1solution of modified question :
the cordinates of any ponit on line as a measure of distance'r' from the point P is
(\frac{r+2\sqrt{3}}{2},\frac{r\sqrt{3}}{2})
now solving this with parabola equation
3r^2-2r-4(2+2\sqrt3)=0
|r_1.r_2|=\frac{4(2+\sqrt3)}{3}
39Nice usage of "parametric" form coolrocks! I was about to post the very same..lol.
1thanks coolrocks pritish and 'dear ' anand @anand: evn i had done the same mistake for Q5. for which the answer was coming as parabola.... neway i have got the answer..its hyperbola
Q6>
106the equilibrium points are x1 and x2
consider x1
if we move towards +ve x-axis, from x1 then F > 0 while if we move towards -ve x-axis then F < 0 implying tat eqbm is unstable
consider x2
If we move towards +ve x-axis, from x2 then F < 0 (i.e. in opposite direction of motion) while if we move towards -ve x-axis then F > 0 implying tat eqbm is stable
1@ AVINAV
COME TO UR THREAD WITH OOOOOOOOOOOOOOPEEEEEEEEEENNNNNNNNNNN EYES
5.hyperbola xy=-2ba
i already ansdwered it in POST no. 32 ......seems u did not read it AND CAME DANCING alonG i got answeer anyways HYPERBOLA LOL!!!!!!!
1
in this q have equated the centripetal force provided to the tension...
for eg.in T1 L=L/4 ,so foe the remaining mass from L/4 to L the tension holds it and provides the net centripetal force for them...
1N=mv2/R + mg cos @
@ is angle b/w radial and vertical direction for any point!
so,cos@ increases till bike reaches from one end to top most point
so,normal reaction increases
39The second reaction in the chemistry question is from p-block elements, and it is the Laboratory method to prepare ammonia : heating ammonium salts in a mildly basic medium.
So basically you have to memorize your reactions.
1Good work Anand aka Venegance....thanks!!!
@sam plz explain ur ans..its right
Pritish the chem question is from Chemoical Equilibrium and hence must have some concept
39All I know is that in HgCl2 the bonding is covalent as opposed to ionic...I don't know how that's a factor for forward equilibrium or backward eq.
106a possible exp. for second rkn maybe that NH3 escapes as a gas
1hmm...so best possible explanation for second one is this one:
dT=w2xp dx
w=angular velocity
p=density[linear ofcourse :P]
x=distance from the pivot
39Possible hai asish...spontaneity of reaction due to gas formation. Entropy increases in forward direction.