Finding Roots of Quadratic
Trending Questions
Q. The minimum distance between the curves x2+y2+4x+16y+66=0 and y2=8x is
- 3√2 units
- 5√2 units
- 4√2−2 units
- 4√2+2 units
Q. A line L:y=mx+3 meets y-axis at E(0, 3) and the arc of the parabola y2=16x, 0≤y≤6 at the point F(x0, y0). The tangent to the parabola at F(x0, y0) intersects the y-axis at G(0, y1). The slope m of the line L is chosen such that the area of the triangle EFG has a local maximum
Match List I with List II and select the correct answer using the code given below the lists :
List IList II (P)m=(1)12(Q)Maximum area of triangle EFG is (2)4(R)y0=(3)2(S)y1=(4)1
Which of the following is correct combination
Match List I with List II and select the correct answer using the code given below the lists :
List IList II (P)m=(1)12(Q)Maximum area of triangle EFG is (2)4(R)y0=(3)2(S)y1=(4)1
Which of the following is correct combination
- (P)→(4), (Q)→(1), (R)→(2), (S)→(3)
- (P)→(3), (Q)→(4), (R)→(1), (S)→(2)
- (P)→(1), (Q)→(3), (R)→(2), (S)→(4)
- (P)→(1), (Q)→(3), (R)→(4), (S)→(2)
Q. From the vertex of the parabola y2=4ax pair of perpendicular chords are drawn. If this chords are adjacent sides of a rectangle, then the locus of the vertex of the rectangle diagonally opposite to the vertex of parabola is
- y2=4a(x−4a)
- y2=4a(x−6a)
- y2=4a(x−2a)
- y2=4a(x−8a)
Q. A line L:y=mx+3 meets y –axis at E(0, 3) and the arc of the parabola y2=16x, 0≤y≤6 at the point F(x0, y0). The tangent to the parabola at F(x0, y0) intersects the y-axis at G(0, y1). The slope m of the line L is chosen such that the area of the triangle EFG has a local maximum.
Match List-I with List-II and select the correct answer using the code given below the lists:
Match List-I with List-II and select the correct answer using the code given below the lists:
List - I | List - II |
P. m = | 1. 12 |
Q. Maximum area of ΔEFG is | 2. 4 |
R. y0= | 3. 2 |
S. y1= | 4. 1 |
- P Q R S
4 1 2 3 - P Q R S
1 3 2 4 - P Q R S
3 4 1 2 - P Q R S
1 3 4 2
Q. The area between the curves y=logex and y=(logex)2 is
- 3−e
- e−loge3
- e−2
- 1
Q. From the vertex of the parabola y2=4ax pair of perpendicular chords are drawn. If this chords are adjacent sides of a rectangle, then the locus of the vertex of the rectangle diagonally opposite to the vertex of parabola is
- y2=4a(x−4a)
- y2=4a(x−6a)
- y2=4a(x−2a)
- y2=4a(x−8a)