RankLab
Back to Questions
MathsMediumMCQ2023 Β· 10 Apr Shift 1

Q66.Let the ellipse 𝐸: π‘₯2 + 9𝑦2 = 9 intersect the positive π‘₯- and 𝑦-axes at the points 𝐴 and 𝐡 respectively. Let the major axis of 𝐸 be a diameter of the circle 𝐢. Let the line passing through 𝐴 and 𝐡 meet the circle 𝐢 at the π‘š point 𝑃. If the area of the triangle with vertices 𝐴, 𝑃 and the origin 𝑂 is 𝑛, where π‘š and 𝑛 are coprime, then π‘š- 𝑛 is equal to (1) 16 (2) 15 (3) 17 (4) 18

What This Question Tests

This question combines properties of ellipses, circles, and straight lines, requiring calculation of intersection points and triangle area.

Concepts Tested

Equation of ellipseEquation of circleLine equationArea of triangle

Formulas Used

x^2/a^2 + y^2/b^2 = 1

Area = 1/2 * |x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|

πŸ“š NCERT Sections This Tests

2.1 β€” Two Charges 5 Γ— 10–8 C And –3 Γ— 10–8 C Are Located 16 Cm Apart. At

Physics Class 11 Β· Chapter 2

70% match

2.1 Two charges 5 Γ— 10–8 C and –3 Γ— 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

2.2 β€” A Regular Hexagon Of Side 10 Cm Has A Charge 5 Mc At Each Of Its

Physics Class 11 Β· Chapter 2

69% match

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.

9.17 β€” (A) Sin IΒ’C = 1.44/1.68 Which Gives IΒ’C = 59Β°. Total Internal Reflection

Physics Class 12 Β· Chapter 9

69% match

9.17 (a) sin iΒ’c = 1.44/1.68 which gives iΒ’c = 59Β°. Total internal reflection takes place when i > 59Β° or when r < rmax = 31Β°. Now, (sin i /sin r max max ) = 1.68 , which gives imax ~ 60Β°. Thus, all incident rays of angles in the range 0 < i < 60Β° will suffer total internal reflections in the pipe. (If the length of the pipe is finite, which it is in practice, there will be a lower limit on i determined by the ratio of the diameter to the length of the pipe.) (b) If there is no outer coating, iΒ’c = sin–1(1/1.68) = 36.5Β°. Now, i = 90Β° will have r = 36.5Β° and iΒ’ = 53.5Β° which is greater than iΒ’c. Thus, all incident rays (in the range 53.5Β° < i < 90Β°) will suffer total internal reflections.