second innings Qs

Thank 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
.

y² - 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...

even 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

5. parabola

3.line is x=-33/16

Arey 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

@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 :)

5.hyperbola xy=-2ba

for Q4
take P=(√3,0)

and it isfind PA.PB

line equation is

y-√3x+3=0

solution 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}

Nice usage of "parametric" form coolrocks! I was about to post the very same..lol.

thanks 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>

the 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

@ 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!!!!!!!

N=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

The 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.

Good 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

All 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.

a possible exp. for second rkn maybe that NH3 escapes as a gas

hmm...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

so tht proves T2>T1

but then the question comes it tht how is sam correct then???

Possible hai asish...spontaneity of reaction due to gas formation. Entropy increases in forward direction.

ans 2> looks to be T₁ > T₂ (AT FIRST SIGHT ) !!!