TargetIIT

1-The point  on the ellipse  x[p]2[/p]+2y[p]2[/p]=6 closest to line x+y=7 is------

2.Eccentricity of ellipse ax[p]2[/p]+bx[p]2[/p]+2fx+2gy+c=0,if major axis
is parallel to -x axis is__
Ans. [sqrt](b-a)/b [/sqrt]

3.Tangents are drawn from the points on the line x-y-5=0,to x[p]2[/p]+4y[p]2[/p]=4,then all the chords of contacts pass through a fixed point whose coordinates are-----?

4.The tangent at point P(cosθ,bsinθ)of a standard ellipse,meets its auxillary circle in two points the chord joining which subtends aright angle  at the centre,then eccentricity is ------?
Ans. (1+sinθ)[p]-1/2[/p]

draw the ellipse from any site and the line they will calculate and give you the shortest distance. simple

This needs to be  done manually,and I am not  sure whether anybody
gave you  some special ticket to use the net during  JEE.

the key here common normal ....

use parametric form of normal which is

√6xsec@-√3ycosec@=3

now it has to perp to line as well => slope is +1

or √2tan@ = 1  

thus co ordinates are .....

√6cos@ , √3sin@

√6(√2/√3) ,√3(1/√3)

2,1 ....


hope ans is correct

and perp dist = |2+1-7|/√2 =2√2...

general eqxn of ellipse || to x axis

(x-h)2/A2 + (y-k)2/B2 = 1

expand&compair coeff with the given eq. find a,b

e=root(1-B2/A2)

if ur nt gettin tell me ill give full soln.

note 'A' and 'a' not same and A2 means A SQUARE

EQUATION OF CHORD OF CONTACT IS T=0 ....

so

hx + 4ky =4  [is eq. of chord of contact frm variable point h,k]

h-k=5.........

or h=k+5

subs. in prev eq.

kx + 5x + 4ky = 4

k(x+y) + 5x - 4 = 0 is of form kL1 + L2 =0

so the common point is the pt of intersection of L1 and L2 is (4/5,-4/5)

tangent at that point is

acos@x + bsin@y = 1 ...[again parametric frm]

now we homogenise it with the auxillary circle x^2 + y^2 = a^2 [actually there are 2 aux circ. im assumin this 1]

so we get

x^2 + y^2 = a^2[acos@x+bsin@y]^2

now since they subtend rt agle ..coeff of x^2 +coeff of y^2 = 0

or

(2)=a^4cos^2@ +a^2b^2sin^@

2= a^4 + a^4sin^2@[b^2/a^2 - 1]

2 = a^4 [1 - e^2 sin^@]

INFO INSUFFICIENT AS a WILL ALSO BE INCLUDED IN ANSWER....

if u find some errors in working let me kno...

Thanks a lot for the soln.....But ans of no.3 has been given to be
(-4/5,-1/5).
It would be very kind of  you to help me out with no 2.(Finding the
eccentricity.....)I,m still not getting it.  

I have solved no 3....
It is as follows...........

Let P(h,h-5),
so from equation of tangent,
xh+4y(h-5)=4,
h(x+4y)-(20y+4)=0
x=4/5,y=-1/5