The equation of the ellipse is
\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\)
\(\therefore \ { a }^{ 2 }=25\ and \ { b }^{ 2 }=9\)
\(\text { Eccentricity of ellipse }=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 9 }{ 25 } } =\frac { 4 }{ 5 } \)
\(\text { So the co-ordinates of foci are } (\pm 4,0)\)
\(\text { Let equation of required hyperbola is}\)
\(\frac { { x }^{ 2 } }{ { a' }^{ 2 } } -\frac { y^{ 2 } }{ { b }'^{ 2 } } =1\)
\(\text { Let e' be the eccentricity of required hyperbola then,}\)
\(a'e'=4\Rightarrow 2e'=4\Rightarrow e'=2\)
\(\therefore b'^{ 2 }={ a' }^{ 2 }({ e' }^{ 2 }-1)\Rightarrow b'^{ 2 }=4(4-1)=12\)
\(\text { Thus equation of required hyperbola is}\)
\(\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 12 } =1\)

Here the line 2x - 5y = 0 passes through point B(1, 2) and intersect 3x - 5y + 8 = 0 at point A.
So the coordinates of point A will be obtained by solving 2x- 5y = 0 and 3x- 5y + 18 = 0.
∴ coordinates of point A are \(\left(-8,{-16\over 5}\right)\)
Now PQ =\(\sqrt{(1+8)^2+\left(2+{16\over5}\right)^2 }=\sqrt{81+{676\over25}}\)
\(=\sqrt{20258+676\over 25}=\sqrt{2701\over25}\)
\(={\sqrt{2701}\over 5}\)units

Given limit \(=\lim_ { x\rightarrow 2 }{ lim } \frac { (x-1)(x-2)(x-4) }{ (x-2)(x+4) } \)
\(-\frac { 1 }{ 3 } \)

Let equation of parabola be
y²=4ax ...(i)
and its vertex is (0,0)
Again, let any point on the parabola be P(h,k)
On putting the values x=h and y=k in Eq(i), we get
 k²=4ah...(ii)
Let OP(=1) be the line segment joining the vertex and point P. Also, it makes an angle \(\theta \) with the X-axis.

\(In\quad right\quad angled\quad \triangle OAP,\)
\(sin\theta =\frac { PA }{ OP } \Rightarrow sin\theta =\frac { k }{ l } \Rightarrow k=lsin\theta \)
\(and\quad cos\theta =\frac { OA }{ OP } \Rightarrow cos\theta =\frac { h }{ l } \Rightarrow h=cos\theta \)
\(On\quad substituting\quad the\quad values\quad of\quad h\quad and\quad k\quad in\quad Eq.(ii),\quad we\quad get\)
\( l^{ 2 }{ sin }^{ 2 }\theta =4alcos\theta \Rightarrow l=\frac { 4acos\theta }{ sin^{ 2 }\theta } \)
\(which\quad is\quad the\quad required\quad length\)

(1 + x)³ (1 + x)⁶
= (1 + x³ + 3x + 3x²) (1 + ⁶C₁ x + ⁶C₂ x² - ⁶C₃​​​ ​​​​x³ + ⁶C₄​​​ ​​​​x⁴ - ⁶C₅​​​ ​​​​x⁵ + ⁶C6​​​ ​​​​x⁶)
Coefficient of x⁵ = - ⁶C₅​​​ + ⁶C₂  + 3 ⁶C₄​​​ - 3 ⁶C₃
= - 6 + 15 + 45 - 60 = - 6

Total number of flags possible = ⁸P₃

Here z= -1+i=r(cos\(\theta \)+i sin\(\theta \))
⇒ rcos \(\theta \)= -1 and r sin\(\theta \)=1   ..(i)
Squaring both sides of (i) and adding
r²(cos²\(\theta \)+sin²\(\theta \)) = 1+1
∴  \(\sqrt { 2 } \)cos \(\theta \)=-1 and \(\sqrt { 2 } \)sin\(\theta \) =1
⇒ cos \(\theta \)=\(-\frac { 1 }{ \sqrt { 2 } } \) and sin\(\theta \) =\(\frac { 1 }{ \sqrt { 2 } } \)
Since sin\(\theta \) is positive and cos\(\theta \) is negative
∴ \(\theta \) lies in second quadrant 
∴  \(\theta =\left( \pi -\frac { \pi }{ 4 } \right) =\frac { 3\pi }{ 4 } \)
Hence polar form of z is 
\(\sqrt { 2 } \left( cos\frac { 3\pi }{ 4 } +i\quad sin\frac { 3\pi }{ 4 } \right) \)

n(A\(\cup \)B)\(\ge \)n(B)=6

The number of people who were liking neither (A) 'Aitraaz' nor 'Veer Zaara' (V) is given by
\(n({ V }^{ ' }\cap { A }^{ , }) ={ n(V\cup A) }^{ ' }\)
=\(n(\cup )-n(V\cup A)\)
=\(n(\cup )-n(V)-n(A)+n(V\cap A)\)

Let the length of the shortest side of the triangle be x cm.
Then length of the longest side =3x cm.
Thus the length of the third side =(3x−2) cm.
Since the perimeter of the triangle is at least 61 cm,
x+3x+(3x−2)  ≥ 61
⇒7x−2≥61
⇒7x≥61+2
⇒7x≥63⇒x≥9 
Thus the minimum length of the shortest side is 9 cm.