# Parabola from 'ARI-HAN't

1)The triangle formed by the tangent to the curve f(x) = x2 + bx - b at the point (1,1) and the co-ordinate axes lie in the first quadrant. If it's area is 2 then the value of b is:

a)-1 b)3 c)-3 d)1

ans c

2)If the normals from any point to the parabola x2 = 4y cuts the line y = 2 in points whose abscissae are in AP , then the slopes of the tangents at the three co-normal points are in:

a) AP b)GP c) HP d) None of these

ans b

3)A line L passing through the focus of the parabola y2 = 4(x-1) intersects the parabola in two distinct points. If 'm' be the slope of the line L then:

a) m E (-1,1) b) m E (-âˆž,-1) U (1 , âˆž) c) m E R d) None of these

ans d

4)The latus rectum of the parabola x = at2 + bt + c and y = a't2 + b't + c' is

a)(aa' - bb')2(a2 + a'2)3/2 b)(ab' - a'b)2(a2 + a'2)3/2 c) (bb' - aa')2(b2 + b'2)3/2 d)(a'b - ab')2(b2 + b'2)3/2

ans b

If the tangents to to the parabola y2 = 4ax at the points (x1 , y1) and (x2 , y2) meet at (x3 , y3) then:

a)y3 = √y1y2 b) 2y3 = y1 + y2 c)2y3 = 1y1 + 1y2 d) None of these.

1 to 4 are doubts (please give the solutions) and 5 I am not getting the answer as given by the book..(someone please confirm which is the correct answer)

11
SANDIPAN CHAKRABORTY ·

http://targetiit.com/iit-jee-forum/posts/parabola-from-ari-han-t-18365.html

21
Shubhodip ·

2)

the equation of a normal to the parabola x^= 4y is m2y=m3x + 2m2+1

(easy to derive)

so from a point h,k three normals can be drawn say with slope m1,m2,m3

so we have m1*m2*m3=-1

and m1*m2 + m2*m3 + m1*m3 = 0
(theory of equation)

or 1/m1 + 1/m2 + 1/m3=0

the x co ordinates can be easily found

as they are in A.P we have

1/m13 + 1/m33 = 2/ m23

using a3 + b3 = (a+b)3 - 3ab(a+b)

u will get 1/m1 * 1/m3 = -(1/m2)

so done...

21
Shubhodip ·

EDIT : u will get 1/m1 * 1/m3 = (1/m2)2

so they are in GP

1
Vinay Arya ·

(1)The given curve is as follows:
F(x)=x2+bx-b
y=x2+bx-b
Differentiating both sides w.r.t. to x.
dydx=2x+b
dydx(1,1)=2+b
The equation of line passing through (1,1) with slope 2+b
y-1=(2+b)(x-1)
y-1=(2+b)x-2-b
(2+b)x-y=1+b
x1+b2+b +y-(1+b)=1
So the coordinates of the triangle are
(0,0),((2+b)1+b,0),(0,-(1+b))
Use triangle formula to find the area.Pluggin these values.You will get b=-3.So (c) is the correct choice.

1
ajoy abcd ·

5.(b)
I used intersection point (at1t2,a(t1+t2))

1
Ashu shah ·

3) any line passing through focus intersects the parabola in 2 distinct points except the axis of parabola so their will be infinite lines passing through focus and cutting parabola in two points so slope(m)=real no except 0 becoz slope 0 means x axis which cannot be included.