Q68.Let the tangent and normal at the point (3√3, 1) on the ellipse x236 and B respectively. Let the circle C be drawn taking AB as a diameter and the line x = 2√5 intersect C at the points P and Q. If the tangents at the points P and Q on the circle intersect at the point (α, β), then α2 −β2 is equal to (1) 61 (2) 60 (3) 304 (4) 314 5 5
What This Question Tests
This is a multi-concept question combining properties of ellipses (tangent, normal) and circles (diameter, tangents, chord of contact) with coordinate geometry.
Concepts Tested
Formulas Used
x x_1/a^2 + y y_1/b^2 = 1 (tangent to ellipse)
(a^2 x)/x_1 - (b^2 y)/y_1 = a^2 - b^2 (normal to ellipse)
(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0 (circle with diameter)
T = 0 (tangent to circle)
Chords of contact: x x_1 + y y_1 = r^2
📚 NCERT Sections This Tests
2.1 — Two Charges 5 × 10–8 C And –3 × 10–8 C Are Located 16 Cm Apart. At
Physics Class 11 · Chapter 2
2.1 Two charges 5 × 10–8 C and –3 × 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.
9.17 — (A) Sin I¢C = 1.44/1.68 Which Gives I¢C = 59°. Total Internal Reflection
Physics Class 12 · Chapter 9
9.17 (a) sin i¢c = 1.44/1.68 which gives i¢c = 59°. Total internal reflection takes place when i > 59° or when r < rmax = 31°. Now, (sin i /sin r max max ) = 1.68 , which gives imax ~ 60°. Thus, all incident rays of angles in the range 0 < i < 60° 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, i¢c = sin–1(1/1.68) = 36.5°. Now, i = 90° will have r = 36.5° and i¢ = 53.5° which is greater than i¢c. Thus, all incident rays (in the range 53.5° < i < 90°) will suffer total internal reflections.
3.10 — In A Reaction Between A And B, The Initial Rate Of Reaction (R0) Was Measured
Chemistry Class 11 · Chapter 3
3.10 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below: A/ mol L–1 0.20 0.20 0.40 B/ mol L–1 0.30 0.10 0.05 r0/mol L–1s–1 5.07 × 10–5 5.07 × 10–5 1.43 × 10–4 What is the order of the reaction with respect to A and B? 3.11 The following results have been obtained during the kinetic studies of the reaction: 2A + B ® C + D Experiment [A]/mol L–1 [B]/mol L–1 Initial rate of formation of D/mol L–1 min–1 I 0.1 0.1 6.0 × 10–3 II 0.3 0.2 7.2 × 10–2 III 0.3 0.4 2.88 × 10–1 IV 0.4 0.1 2.40 × 10–2 Determine the rate law and the rate constant for the reaction. 3.12 The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/ mol L–1 [B]/ mol L–1 Initial rate/ mol L–1 min–1 I 0.1 0.1 2.0 × 10–2 II – 0.2 4.0 × 10–2 III 0.4 0.4 – IV – 0.2 2.0 × 10–2 3.13 Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s–1 (ii) 2 min–1 (iii) 4 years–1 3.14 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample. 3.15 The experimental data for decomposition of N2O5 [2N2O5 ® 4NO2 + O2] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 102 × [N2O5]/ 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 mol L–1 (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N2O5] and t. (iv) What is the rate law ? Chemistry 86 Reprint 2025-26 (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).
📋 Question Details
- Chapter
- Ellipse
- Topic
- Tangent, Normal, and Circle properties
- Year
- 2023
- Shift
- 13 Apr Shift 1
- Q Number
- Q68
- Type
- MCQ
- NCERT Ref
- Class 11 Mathematics Ch 11: Conic Sections | Class 11 Mathematics Ch 10: Straight Lines | Class 11 Mathematics Ch 12: Circles (Implicitly)
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