EDIT : u will get 1/m1 * 1/m3 = (1/m2)^{2}
so they are in GP
1)The triangle formed by the tangent to the curve f(x) = x^{2} + 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 x^{2} = 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 y^{2} = 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 = at^{2} + bt + c and y = a't^{2} + b't + c' is
a)(aa' - bb')^{2}(a^{2} + a'^{2})^{3/2} b)(ab' - a'b)^{2}(a^{2} + a'^{2})^{3/2} c) (bb' - aa')^{2}(b^{2} + b'^{2})^{3/2} d)(a'b - ab')^{2}(b^{2} + b'^{2})^{3/2}
ans b
5)answer not matching
If the tangents to to the parabola y^{2} = 4ax at the points (x_{1} , y_{1}) and (x_{2} , y_{2}) meet at (x_{3} , y_{3}) then:
a)y^{3} = √y_{1}y_{2} b) 2y_{3} = y_{1} + y_{2} c)2y_{3} = 1y_{1} + 1y_{2} d) None of these.
book answer a my answer b
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)
sorry two similar threads has been made...
please refer to
http://targetiit.com/iit-jee-forum/posts/parabola-from-ari-han-t-18365.html
2)
the equation of a normal to the parabola x^= 4y is m^{2}y=m^{3}x + 2m^{2}+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/m1^{3} + 1/m3^{3} = 2/ m2^{3}
using a^{3} + b^{3} = (a+b)^{3} - 3ab(a+b)
u will get 1/m1 * 1/m3 = -(1/m2)
so done...
(1)The given curve is as follows:
F(x)=x^{2}+bx-b
y=x^{2}+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.
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.