Maths

Given hyperbola :

$\frac{x^2}{a^2}-\frac{y^2}{9}=1$

$$ $?$  it passes through

$\left(8,3\sqrt[]{3}\right)$

$?\frac{64}{a^2}-\frac{27}{9}=1\Rightarrow a^2=16$

Now, equation of normal to hyperbola

$\frac{16x}{8}+\frac{9y}{3\sqrt[]{3}}=16+9$

$\text{exception 25:}$ $\left(-1,9\sqrt[]{3}\right)$  satisfied

  $E:\frac{x^2}{4}+\frac{y^2}{2}=1$

any pt on it is P  $\left(2cos\theta ,\sqrt[]{2}sin\theta \right)$

$\therefore$ M (h, k) be mid point of P & A (4, 3)

$\Rightarrow {\left(h-2\right)}^2+{\left(\frac{2k-3}{\sqrt[]{2}}\right)}^2=1$        

Required locus (x – 2)² +   $\frac{{\left(y-\frac{3}{2}\right)}^2}{\frac{1}{2}}=1$

$\therefore e=\sqrt[]{1-\frac{1}{2}}=\frac{1}{\sqrt[]{2}}$

Equation of any tangent to   $\frac{x^2}{16}+\frac{y^2}{9}=1$

is y = mx + $\sqrt[]{16m^2+9}$  if this line is also tangent to x² + y² = 12

then  $\sqrt[]{12}$ =   $\left|\frac{\sqrt[]{16m^2+9}}{\sqrt[]{1+m^2}}\right|$

^{$\Rightarrow 12+12m^2=16m^2+9$}

$\Rightarrow 12m^2=9$        

  $?\frac{dy}{dx}+e^x\left(x^2-2\right)y=\left(x^2-2x\right)\left(x^2-2\right)e^{2x}$ $\frac{}{}{}^{}{}^{}{}^{}{}^{}{}^{}$

Here, I.F.

$=e^{{\displaystyle \int e^x\left(x^2-2\right)dx}}$

$=e^{\left(x^2-2x\right)e^x}$

$$ $\therefore$  Solution of the differential equation is

$y.e^{\left(x^2-2x\right)e^x}=\left(x^2-2x-1\right)e^{\left(x^2-2x\right)e^x}$

$$ $\therefore$  y(0) = 0

$\therefore c=1$

$\therefore y\left(2\right)=-1+1=0$

RM = $\left|\frac{3+7-5}{\sqrt[]{2}}\right|=\frac{5}{\sqrt[]{2}}$

$\mathcal{l}sin60°=\frac{5}{\sqrt[]{2}}\Rightarrow \mathcal{l}=5\sqrt[]{\frac{2}{3}}$

$\therefore Areaof\Delta PQR=\frac{\text{exception 25:}}{4}{\mathcal{l}}^2=$=25/2√3

  $x\frac{dy}{dx}+2y=x{e}^x$ $\frac{}{}{}^{}$

$\frac{dy}{dx}+\frac{2y}{x}=e^x$

$\frac{dz}{dx}=e^x.2\left(x-1\right)+e^x{\left(x-1\right)}^2=0$

$e^x\left(x-1\right)\left(2+x-1\right)=0$

$x=-1,1$

x = 1 local maxima. Then maximum value is

$z\left(-1\right)=\frac{4}{e}-e$

$\left|x^2-9\right|=3$

$\Rightarrow x=\pm 2\sqrt[]{3},\pm \sqrt[]{6}$

Required area = A

$\frac{A}{2}={\displaystyle \underset{0}{\overset{\sqrt[]{6}}{\int }}\left(9-x^2-3\right)dx+{\displaystyle \underset{0}{\overset{3}{\int }}\left(\sqrt[]{9+y}-\sqrt[]{9-y}\right)dy}}$

$A=16\sqrt[]{6}+32\sqrt[]{3}-72=8\left[2\sqrt[]{6}+4\sqrt[]{3}-9\right]$

Note : No option in the question paper is correct.

Put x = cos 2θ

dx = -2 sin 2θ . dθ

$={\displaystyle \int \frac{1}{cos2\theta }tan\theta \left(-4sin\theta .cos\theta \right)d\theta }$

$g\left(\frac{1}{2}\right)=\mathcal{l}n\left|2-\sqrt[]{3}\right|+\frac{\pi }{3}$

= $\mathcal{l}n\left|\frac{\sqrt[]{3}-1}{\sqrt[]{3}+1}\right|+\frac{\pi }{3}$  

Let p₁ : y² = 8x

p₂ : y² = 16 (3 – x) = -16 (x – 3)

finding their intersection points.

y² = 8x & y² = -16 (x – 3)

8x = -16x + 48

$=2.{\displaystyle \underset{0}{\overset{4}{\int }}\left(3-\frac{y^2}{16}-\frac{y^2}{8}\right)dy}$

$=2{\left(3y-\frac{y^3}{3*16}-\frac{y^3}{3*8}\right)}_0^4$ = 16

$?s_1+s_2=k$                

76x² + 3πr² = k

$\therefore 152x\frac{dx}{dr}+6\pi r=0$

Now

$V=40x^3+\frac{2}{3}\pi r^3$

$\Rightarrow \left(\frac{x}{r}\right)=\frac{152}{3}\frac{1}{120}=\frac{19}{45}$