Q7. Let the parabola y = x2 + px −3, meet the coordinate axes at the points P, Q and R . If the circle C with centre at (−1, −1) passes through the points P, Q and R, then the area of △PQR is : (1) 7 (2) 4 (3) 6 (4) 5
What This Question Tests
This question combines concepts of parabola's intercepts, circle's equation passing through points, and calculates the area of a triangle formed by these intercepts.
Concepts Tested
Formulas Used
x-intercept: y=0
y-intercept: x=0
Area of triangle = 1/2 |x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|
Equation of circle x^2+y^2+2gx+2fy+c=0
📚 NCERT Sections This Tests
2.2 — A Regular Hexagon Of Side 10 Cm Has A Charge 5 Mc At Each Of Its
Physics Class 11 · Chapter 2
2.2 A regular hexagon of side 10 cm has a charge 5 mC at each of its vertices. Calculate the potential at the centre of the hexagon.
12.7 — The Radius Of The Innermost Electron Orbit Of A Hydrogen Atom Is
Physics Class 12 · Chapter 12
12.7 The radius of the innermost electron orbit of a hydrogen atom is 5.3×10–11 m. What are the radii of the n = 2 and n =3 orbits?
6.11 — Dynamics Of Rotational
Physics Class 11 · Chapter 6
6.11 Dynamics of rotational the motion of extended bodies. motion about a fixed axis A large class of problems with extended bodies can be
📋 Question Details
- Chapter
- Coordinate Geometry
- Topic
- Circles and Parabolas
- Year
- 2025
- Shift
- 22 Jan Shift 1
- Q Number
- Q7
- Type
- MCQ
- NCERT Ref
- Class 11 Mathematics Ch 10: Straight Lines; Class 11 Mathematics Ch 11: Conic Sections (Parabola & Circle)
More from this Chapter
Q78.The perpendicular bisector of the line segment joining P(1, 4) and Q(k, 3) has y-intercept - 4. Then a possible value of k is (1) 1 (2) 2 (3) −2 (4) −4
Q72.The normal at (2, 23 ) to the ellipse, x216 + y23 = 1 touches a parabola, whose equation is (1) y2 = −104x (2) y2 = 14x (3) y2 = 26x (4) y2 = −14x sin(π cos2 x)
Q68.A light ray emerging from the point source placed at P(1, 3) is reflected at a point Q in the axis of x. If the reflected ray passes through the point R (6, 7), then the abscissa of Q is: (1) 1 (2) 3 (3) 7 (4) 5 2 2
Q76.If two vertices of an equilateral triangle are A(−a, 0) and B(a, 0), a > 0, and the third vertex C lies above x- axis then the equation of the circumcircle of △ABC is : (1) 3x2 + 3y2 −2√3ay = 3a2 (2) 3x2 + 3y2 −2ay = 3a2 (3) x2 + y2 −2ay = a2 (4) x2 + y2 −√3ay = a2