1. Scenario-1: 5x2+9y2−30x=0 5 x^{2}+9 y^{2}-30 x=0 5x2+9y2−30x=0.
Scenario-2: (8,3) (8,3) (8,3) and (16,3) (16,3) (16,3) are foci of a hyperbola and eccentricity is 4 .
Find the equation of a hyperbola whose parametric coordinates is (3secθ,2tanθ) (\sqrt{3} \sec \theta, 2 \tan \theta) (3secθ,2tanθ).
From scenario-1, show that the equation indicates and ellipse, find the equation of its latus rectum.
From scenatio-2, find the equation of the hyperbola.
2. Scenario-1: f(x)=x4−3x3−11x2+23x−10 f(x)=x^{4}-3 x^{3}-11 x^{2}+23 x-10 f(x)=x4−3x3−11x2+23x−10
Scenario-2: g(x)=x3−3x2−8x+30 g(x)=x^{3}-3 x^{2}-8 x+30 g(x)=x3−3x2−8x+30.
If α,β \alpha, \beta α,β are two roots of the equation x2+5x+3= x^{2}+5 x+3= x2+5x+3= 0 , then find the value of 1β−1α \frac{1}{\beta}-\frac{1}{\alpha} β1−α1.
From scenario-2, if 3+i 3+i 3+i is a root of the equation g(x)=0 g(x)=0 g(x)=0, find the other roots.
From scenario-1, if 1 is a root of the equation f(x) f(x) f(x) =0 =0 =0 and other roots are α,β,γ \alpha, \beta, \gamma α,β,γ, then find α3+β3+ \alpha^{3}+\beta^{3}+ α3+β3+ r3 r^{3} r3.
3. Scenario: f(x)=ax2+bx+c,a≠0 f(x)=a x^{2}+b x+c, a \neq 0 f(x)=ax2+bx+c,a=0 is a quadratic function.
If a=1,b=−2,c=1 a=1, b=-2, c=1 a=1,b=−2,c=1, then find the nature of the roots of the equation f(x)=0 f(x)=0 f(x)=0.
From the scenario, if α,β \alpha, \beta α,β are the roots of the equation f(x)=0 f(x)=0 f(x)=0, then express the roots of the equation cx2− c x^{2}- cx2− (b2a−2c)x+c=0 \left(\frac{b^{2}}{a}-2 c\right) x+c=0 (ab2−2c)x+c=0 in terms of α,β \alpha, \beta α,β.
From scenario if a=1,b=−2n,c=n2−m2 a=1, b=-2 n, c=n^{2}-m^{2} a=1,b=−2n,c=n2−m2 then form an equation whose roots are the sum and absolute value of the difference of the roots of the equation f(x)=0 f(x)=0 f(x)=0.
4. Scenario-1:
Scenario-2:
If two equal forces of magnitude F F F acting at a point at an angle 120∘ 120^{\circ} 120∘, determine the magnitude and direction of their resultant.
How much equal forces of same magnitudes are to be added to each of the forces in scenario-1, so that the point of action of their resultant moves aside 5 cm 5 \mathrm{~cm} 5 cm ?
From scenario-2 if Q=13 N Q=13 \mathrm{~N} Q=13 N and R=12 N R=12 \mathrm{~N} R=12 N is the resultant of P P P and Q Q Q, find the value of P P P.
5. N=tan−1(cosectan−1x−tancot−1x) and f(θ)=cosθ \mathrm{N}=\tan ^{-1}\left(\operatorname{cosec} \tan ^{-1} x-\tan \cot ^{-1} x\right) \text { and } \mathrm{f}(\theta)=\cos \theta N=tan−1(cosectan−1x−tancot−1x) and f(θ)=cosθ
If x=12cos−134 x=\frac{1}{2} \cos ^{-1} \frac{3}{4} x=21cos−143, then what is the value of tanx \tan x tanx.
Show that, N=12tan−1x N=\frac{1}{2} \tan ^{-1} x N=21tan−1x.
Solve: f(θ)+f(2θ)+f(3θ)=0 f(\theta)+f(2 \theta)+f(3 \theta)=0 f(θ)+f(2θ)+f(3θ)=0, when −2π≤θ≤ -2 \pi \leq \theta \leq −2π≤θ≤ 2π 2 \pi 2π.