This needs to be done manually,and I am not sure whether anybody
gave you some special ticket to use the net during JEE.
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
draw the ellipse from any site and the line they will calculate and give you the shortest distance. simple
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