1draw the ellipse from any site and the line they will calculate and give you the shortest distance. simple
1-The point on the ellipse x2+2y2=6 closest to line x+y=7 is------
2.Eccentricity of ellipse ax2+bx2+2fx+2gy+c=0,if major axis
is parallel to -x axis is__
Ans. √(b-a)/b
3.Tangents are drawn from the points on the line x-y-5=0,to x2+4y2=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)-1/2
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8 Answers
11This needs to be done manually,and I am not sure whether anybody
gave you some special ticket to use the net during JEE.
3the 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...
3general 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
3EQUATION 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)
3tangent 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...
1Thanks 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.
1I 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