The coordinates of the point on the parabola y2 = 8x , which is at mininmum distance from circle
x2 + (y-6)2 = 1 are
(A) (2,-4)
(B) (18,12)
(C) (2,4)
(D) none
42) The condition for the two tangents , drawn from a point not lying in the y-axis to the parabola y2 = 4ax , to become normals to the parabola x2 = 4by is that
(A) b>a
(B) ab>4
(C) b2 > 4a2
(D) a2 > 8b2
43) The triangle formed by the tangents to the parabola y2 = 4ax , at the ends of its latus rectum , and the double ordinate through its focus , is
(A)Equilateral
(B) Acute angled isosceles
(C) Right angled isosceles
(D) dependent on the value of a for its classification
211) use this fact tat shortest distance between two curve is always along the common normal of the 2 curves......find the eq of normal through parametric cordinates on parabola(2t2,4t) and since this line shud also be the normal of the circel it shud satisfy cordinates of center of circle(0,6)....using this find t and hence cordinate
ans is (2,4) if i hav not made any calc error......
19i could not understand qs 3) ....a double ordinate through the focus is the latum rectum itself ?????...
4@Decotosh
The qs means the triangle formed by the tgts and latus rectum
12) a2>8b2
write the equation of tangent to y2=4ax in term of slope and euation of normal to x2=4by.......since these two equation are same compare the coeficient of x,y and contant term.
you will get the euation 2bm2-am+b=0
utilize the fact dat discriminant of equation cannnot be <0
4Thanks Archana
@Soumik : I'm not sure of the ans but I think it must be b or c
Let's wait for some more conformations
19well, i misunderstood the question,lol
ans 3> (c) ....we know that the tangents at the extremities of the focal chord of a parabola intersect at right angles on its directrix..hence the answer !
.....proof is easy...take 2 pts P , Q on the extremity of the focal chord ...now take the eqn of the tangents at those pts and see their slopes ! you get something on the directrix as the intersection pt !
