1. Scenario-1: The focus of parabola is S(1,−2) S(1,-2) S(1,−2) and the line 2x−y 2 x-y 2x−y +4=0 +4=0 +4=0 is the tangent of vertex point.
Scenario-2: The coordinates of focus of an ellipse are S(−2 S(-2 S(−2, 0) 0) 0) and S′(2,0) S^{\prime}(2,0) S′(2,0).
Find the eccentricity of hyperbola x29−y216=1 \frac{x^{2}}{9}-\frac{y^{2}}{16}=1 9x2−16y2=1.
Find the equation of the parabola according to scenario-1.
Find the equation of an ellipse that the point (4,0) (4,0) (4,0) lies on an ellipse according to scenario-2.
2. Scenario-1: x=ay2+by+c x=a y^{2}+b y+c x=ay2+by+c
Scenario-2: The coordinates of foci of the hyperbola are S(−6,0) S(-6,0) S(−6,0) and S′(6,0) S^{\prime}(6,0) S′(6,0).
If the ellipse x2p+y225=1 \frac{x^{2}}{p}+\frac{y^{2}}{25}=1 px2+25y2=1 passes through the point (6,4) (6,4) (6,4), find the length of major axis of the ellipse.
Vertex of parabola is (1,2) (1,2) (1,2) and the parabola passes through the point (3,−2) (3,-2) (3,−2), find the values of a,b,c \mathrm{a}, \mathrm{b}, \mathrm{c} a,b,c according to scenario-1.
If the length of latus rectum of hyperbola is 10 units then find the equation of hyperbola according to scenario-2.
3. φ(x)=ax3+bx2+cx+dψ(x)=x2−mx+l. \begin{array}{l} \varphi(\mathrm{x})=\mathrm{ax}^{3}+\mathrm{bx}^{2}+\mathrm{cx}+\mathrm{d} \\ \psi(\mathrm{x})=\mathrm{x}^{2}-\mathrm{mx}+l . \end{array} φ(x)=ax3+bx2+cx+dψ(x)=x2−mx+l.
What will be the value of a when the roots of the equation (a−1)x2−(a+2)x+4=0 (a-1) x^{2}-(a+2) x+4=0 (a−1)x2−(a+2)x+4=0 are real and equal?
If a=4,b=−2,c=0 a=4, b=-2, c=0 a=4,b=−2,c=0 and d=3 d=3 d=3 of the equation φ(x)=0 \varphi(x)=0 φ(x)=0 and the roots are α,β,γ \alpha, \beta, \gamma α,β,γ then find the value of ∑α2β \sum \alpha^{2} \beta ∑α2β.
If a=0,b=1,c=−l a=0, b=1, c=-l a=0,b=1,c=−l and d=m d=m d=m of the equation φ(x)=0 \varphi(x)=0 φ(x)=0. If the roots of the equation φ(x)=0 \varphi(x)=0 φ(x)=0 and ψ(x)=0 \psi(x)=0 ψ(x)=0 differ only by a constant then prove that, l+m+4=0 l+\mathrm{m}+4=0 l+m+4=0.
4. f(x)=cosx;h(x)=tan−1x f(x)=\cos x ; h(x)=\tan ^{-1} x f(x)=cosx;h(x)=tan−1x
Prove that, cot−1(tan2φ)+cot−1(−tan3φ)=φ \cot ^{-1}(\tan 2 \varphi)+\cot ^{-1}(-\tan 3 \varphi)=\varphi cot−1(tan2φ)+cot−1(−tan3φ)=φ.
Solve that, (2+3)f(2θ)=1−f(π2−2θ) (2+\sqrt{3}) \mathrm{f}(2 \theta)=1-\mathrm{f}\left(\frac{\pi}{2}-2 \theta\right) (2+3)f(2θ)=1−f(2π−2θ) according to stem.
Prove that 2h(a−ba+btanθ2)=cos−1b+af(θ)a+bf(θ) 2 h\left(\frac{\sqrt{a-b}}{\sqrt{a+b}} \tan ^{\frac{\theta}{2}}\right)=\cos ^{-1} \frac{b+a f(\theta)}{a+b f(\theta)} 2h(a+ba−btan2θ)=cos−1a+bf(θ)b+af(θ).
5.
Define the resolved part of a force.
According to scenario-1; O \mathrm{O} O is the incenter of triangle ABC A B C ABC and three forces acting are in equilibrium. Show that,
P12:P22:P32=(1+cosA):(1+cosB):(1+cos P_{1}{ }^{2}: P_{2}{ }^{2}: P_{3}{ }^{2}=(1+\cos A):(1+\cos B):(1+\cos P12:P22:P32=(1+cosA):(1+cosB):(1+cos C).
According to scenario-2 how much equal force is added with each of the two forces, the new resultant moves 8 cm 8 \mathrm{~cm} 8 cm away from the old resultant?