If `(a , b)` is the midpoint of a chord passing through the vertex of the parabola `y^2=4x ,` then `a=2b` (b) `a^2=2b` `a^2=2b` (d) `2a=b^2`

IF (a,b) is the mid point of chord passing through the vertex of the parabola y^2=4x , then

If (a,b) is the midpoint of a chord passing through the vertex of the parabola y^(2)=4x then (A) a=2b(B)2a=b(C)a^(2)=2b(D)2a=b^(2)

If (a,b) is the midpoint of a chord passing through the vertex of the parabola y^(2)=4(x+1), then prove that 2(a+1)=b^(2)

The vertex of the parabola y^(2) =(a-b)(x -a) is

Let P be the mid-point of a chord joining the vertex of the parabola y^(2)=8x to another point on it.Then,the locus of P is (A)y^(2)=2x (B) y^(2)=4x(C)(y^(2))/(4)+y^(2)=1(D)x^(2)+(y^(2))/(4)=1

Two thin rods AB & CD of lengths 2a & 2b move along OX & OY respectively, when 'O' is the origin. The equation of locus of the centre of the circle passing through the extremeties of the two rods is: a. x^2+y^2=a^2+b^2 b. x^2-y^2=a^2-b^2 c. x^2+y^2=a^2-b^2 d. x^2-y^2=a^2+b^2

The vertex of the parabola (y-2)^(2)=16(x-1) is (1,2) b.(-1,2)c(1,-2)d.(2,1)

If a!=0 and the line 2bx+3cy+4d=0 passes through the points of intersection of the parabolas y^(2)=4ax and x^(2)=4ay, then: d^(2)+(2b+3c)^(2)=0 (b) d^(2)+(3b+2c)^(2)=0d^(2)+(2b-3c)^(2)=0 (d) d^(2)+(3b-2c)^(2)=0

If a!=0 and the line 2bx+3cy+4d=0 passes through the points of intersection of the parabolas y^(2)=4ax and x^(2)=4ay, then of (a)d^(2)+(2b+3c)^(2)=0(b)d^(2)+(3b+2c)^(2)=0(c)d^(2)+(2b-3c)^(2)=0 (d)none of these