The length of the tangent to the circle \(x^{2}+y^{2}+2 g x+2 f y+c=0\) from the point 
\(\left(x_{1}, y_{1}\right) \text { is } \sqrt{x_{1}^{2}+y_{1}^{2}+2 g x_{1}+2 f y_{1}+c}\)
Length of the tangent to the given circle
\( =\sqrt{x_{1}^{2}+y_{1}^{2}-4 x_{1}-3 y_{1}+12} \)
\(=\sqrt{2^{2}+3^{2}-4.2-3.3+12} \)
\(=\sqrt{4+9-8-9+12} \)
\(=\sqrt{8} \)
\(=2 \sqrt{2} \text { units } \)

We have,
f(x) = (x-2)², then
\(f'(x)=2(x-2)=0 \) gives x = 2.
The intervals of monotonicity are \((-\infty,2)\) and \((2,\infty)\)
Since  \(f'(x)<0, \forall x \in (-\infty,2)\) for \((-\infty,2)\) the f(x) is strictly decreasing on \((2,\infty)\)
As \(f'(x)>0, \) for \(x \in(2, \infty)\) the function f(x) is strictly increasing on \((2,\infty)\)
Becasue f'(x) changes its sign from negative to positive when passing through x = 2 for the function f(x) it has a local minimum at x = 2
The local minimum value is f(2) = 0.

Given f(x) = tan x, x ∈ [0, π]
Rolle's theorem is not applicable since tan x is not continuous at x = \(\frac{\pi}{2}\) [∵ tan \(\frac{\pi}{2}\) = ∞]

Given y = x⁴ + 2x² - x
\(\frac { dy }{ dx } \) = 4x³ + 4x - 1
Slope of the tangent at x = 1 is
m = \(\left( \frac { dy }{ dx } \right) \)(x = 1)
= 4(1)³+ 4 (1) - 1
= 4+4-1 = 7
∴ m = 7

Given v = x³
Differentiating with respect to x we get,
\(\frac { dv }{ dt } \) = 3x²
When x = 5, \(\frac { dv }{ dt } \) = 3(52) = 75
∴ \(\frac { dv }{ dt } \) when x = 5 is 75 units.

Δ = \(\left| \begin{matrix} 6 & -7 \\ 9 & -5 \end{matrix} \right| \) = -30 + 63 = 33
Δ₁ = \(\left| \begin{matrix} 16 & -7 \\ 35 & -5 \end{matrix} \right| \) = -80 + 245 = 165
Δ₂ = \(\left| \begin{matrix} 6 & 16 \\ 9 & 35 \end{matrix} \right| \) = 210 - 144 = 66
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 165 }{ 33 } \) = 5
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 66 }{ 33 } \) = 2
∴ Solution set is { 5, 2}

Equation of the circle is
2x² + 2y² - 6x + 4y + 2 = 0
Dividing by 2, we get
x² + y² - 3x + 2y + 1 = 0
Here 2g = -3 ⇒ g = \(\frac { -3 }{ 2 } \)
2f = 2 ⇒ f = 1
and c = 1
∴ Centre is (-g, -f) = \(\left( \frac { 3 }{ 2 } ,-1 \right) \)
and r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \) = \(\sqrt { { \left( \frac { 3 }{ 2 } \right) }^{ 2 }+{ 1 }^{ 2 }-1 } \)
= \(\sqrt { \frac { 9 }{ 4 } } =\frac { 3 }{ 2 } \) units.

Let A = \(\left[ \begin{matrix} -2 & 2 & -1 \\ 0 & 5 & 1 \\ 0 & 0 & 0 \end{matrix} \right] \). Then A is a matrix of order 3 × 3 and ρ(A) ≤ 3.
The only third order minor is |A| = \(\left| \begin{matrix} -2 & 2 & -1 \\ 0 & 5 & 1 \\ 0 & 0 & 0 \end{matrix} \right| \) = (-2)(5)(0) = 0. So ρ(A) ≤ 2.
There are several second order minors. We find that there is a second order minor, for example, \(\left| \begin{matrix} -2 & 2 \\ 0 & 5 \end{matrix} \right| \) = (-2)(5) = -10 ≠ 0. So, ρ(A) = 2.
Note that there are two non-zero rows. The third row is a zero row.

\(\left[ \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right] \)
A is a matrix of order 2 \(\times\) 2
∴ \(\rho \)(A) ≤ min(2,2) = 2
The highest order of minor of A is 2
it is \(\left| \begin{matrix} 2 & -1 \\ -1 & 2 \end{matrix} \right| \)= 4 - 4 = 0
So, \(\rho \)(A)<2
Next consider the minor of order 1 |2| = 2 ≠ 0
∴ \(\rho \)(A) = 1

Equation of the circle with end points of the diameter as (x₁, y₁₎ and (x₂, y₂) given in theorem is
(x−x₁)(x−x₂)+(y−y₁)(y−y₂) = 0
(x+4)(x−1)+(y+2)(y−1) = 0
x² + y² + 3x + y − 6 = 0 which is the required equation of the circle.

Since \(f'(x)=1-cosx\ge0\) and zero at the points \(x=2n\pi, n\in Z\) and hence the function is increasing on the real line.
Since there is no sign change in f′(x) when passing through \(x=2n\pi, n\in Z\) by the first derivative test there is no local extrema.

Meclaurin's expansion
\(f(x)=f(0)+\frac { { f }^{ 1 }(0) }{ 1! } x+\frac { { f }^{ II }(0) }{ 2! } { x }^{ 2 }+\frac { { f }^{ III }(0) }{ 3! } { x }^{ 3 }+\).................
log(1-x) = \(0-\frac { 1 }{ 1! } x-\frac { { x }^{ 2 } }{ 2! }- \frac { 2 }{ 3! } { x }^{ 3 }-\frac { { 6x }^{ 4 } }{ 4! } +.....\)
= \(-x-\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 4 } }{ 4 } \)+ .....
log(1 - x) = \(-\left( x+\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 4 } }{ 4 } +.... \right) \)

\(f'\left( x \right) =6\times \frac { 4 }{ 3 } { x }^{ \frac { 4 }{ 3 } -1 }-3\times \frac { 1 }{ 3 } { x }^{ \frac { 1 }{ 3 } -1 }\) 
= \({ 8x }^{ \frac { 1 }{ 3 } }-x^{ \frac { -2 }{ 3 } }\) 
f'(x) = 0
\(\Rightarrow{ 8 }x^{ \frac { 1 }{ 3 } }-\frac { 1 }{ { x }^{ \frac { 2 }{ 3 } } } =0\) 
\(\Rightarrow \frac { 8x-1 }{ { x }^{ \frac { 2 }{ 3 } } } =0\) 
\(\Rightarrow x=\frac { 1 }{ 8 } \)
Thus, the critical number is \(x=\frac { 1 }{ 8 } \) 
Evaluating f(x) at the end points = -1
x = 1 and at the critical number x = \(\frac { 1 }{ 8 } \) 
we get
\(f(-1)=6(-1)^{ \frac { 4 }{ 3 } }-3\left( -1 \right) ^{ \frac { 1 }{ 3 } }\) 
= 6( 1) - 3 (-1) = 6 + 3 = 9
\(f(1)=6(1)^{ \frac { 4 }{ 3 } }-3(1)^{ \frac { 1 }{ 3 } }=6-3=3\)
\(f\left( \frac { 1 }{ 8 } \right) =6\left( \frac { 1 }{ 8 } \right) ^{ \frac { 4 }{ 3 } }-3\left( \frac { 1 }{ 8 } \right) ^{ \frac { 1 }{ 3 } }\) 
=\(6\left( { 2 }^{ -3 } \right) ^{ \frac { 4 }{ 3 } }-3\left( 2^{ -3 } \right) ^{ \frac { 1 }{ 3 } }\) 
= \(\frac { 6 }{ 16 } -\frac { 3 }{ 2 } =\frac { 3 }{ 8 } -\frac { 3 }{ 2 } \) 
= \(\frac { 3-12 }{ 8 } =\frac { -9 }{ 8 } \)
From these values, the absolute maximum is 9 which occurs at x = -1 and the absolute minimum is \(-\frac { 9 }{ 8 } \) which occurs at x = \(\frac { 1 }{ 8 } \)

Equating of the given curve is y = x⁴+ 2e^x
\(\frac { dy }{ dx } \) =  x⁴+ 2e^x
m = \(\frac { dy }{ dx } \)_{(0, 2)}
= 4(0) + 2e⁰ = 2(1) = 2
∴ Equating of the tangent is y - y₁ = m (x - x₁)
⇒ y - 2 = 2 (x - 0)
⇒ y-2 =2x
⇒ 2x - y = -2
Equating of the normal is y -y₁ = \(\frac{-1}{m}\) (x-x₁)
⇒ y - 2 = -\(\frac12\) (x - 0)
⇒ 2y - 4 = -x
⇒ x + 2y - 4 = 0

The latus rectum LL' of the hyperbola passes through S(ae, 0)

∴ L is (ae, y₁)
Substituting L in \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } \) = 1 we get,
Substituting L in \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\frac { { a }^{ 2 }{ e }^{ 2 } }{ { a }^{ 2 } } -\frac { { { y }_{ 1 } }^{ 2 } }{ { b }^{ 2 } } =1\Rightarrow { e }^{ 2 }-\frac { { { y }_{ 1 } }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\Rightarrow { e }^{ 2 }-1=\frac { { { y }_{ 1 } }^{ 2 } }{ { b }^{ 2 } } \) \([{ b }^{ 2 }={ a }^{ 2 }({ e }^{ 2 }-1)\)
\(\Rightarrow { { y }_{ 1 } }^{ 2 }={ b }^{ 2 }({ e }^{ 2 }-1)\) \(\Rightarrow \frac { { b }^{ 2 } }{ { a }^{ 2 } } ={ e }^{ 2 }-1]\)
\(\Rightarrow { { y }_{ 1 } }^{ 2 }={ b }^{ 2 }\left( \frac { { b }^{ 2 } }{ { a }^{ 2 } } \right) \) \(\Rightarrow { { y }_{ 1 } }^{ 2 }=\frac { { b }^{ 4 } }{ { a }^{ 2 } } \)
\(\Rightarrow { y }_{ 1 }=\pm \frac { { b }^{ 2 } }{ a } \)
∴ End points oflatus rectum Land L' are
\(\left( ae,\frac { { b }^{ 2 } }{ a } \right) \) and \(\left( ae,-\frac { { b }^{ 2 } }{ a } \right) \)
Hence, the length of latus rectum LL' = \(\frac { { b }^{ 2 } }{ a } +\frac { { b }^{ 2 } }{ a } =\frac { 2{ b }^{ 2 } }{ a } \) units.
Hence proved.

2x+5y = -2, x+2y = -3
The matrix form of the system is
\(\left( \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} -2 \\ -3 \end{matrix} \right) \)
⇒ AX = B where
A =\(\left( \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right) ,B=\left( \begin{matrix} -2 \\ -3 \end{matrix} \right) \)
X =\(\left( \begin{matrix} x \\ y \end{matrix} \right) \)
⇒ = A^-1B
|A| = \(\left| \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right| \)= 4 - 5 = -1 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adjA=\frac { 1 }{ -1 } \left[ \begin{matrix} 2 & -5 \\ -1 & 2 \end{matrix} \right] \)
=\(\left[ \begin{matrix} -2 & 5 \\ 1 & -2 \end{matrix} \right] \)
∴ X = A^-1B =\(\left[ \begin{matrix} -2 & 5 \\ 1 & -2 \end{matrix} \right] \left[ \begin{matrix} -2 \\ -3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 4-15 \\ -2+6 \end{matrix} \right] =\left[ \begin{matrix} -11 \\ 4 \end{matrix} \right] \)
∴ Solution set is x = -11, y = 4

\(\left[ \begin{matrix} 3 & -1 & 2 \\ -6 & 2 & 4 \\ -3 & 1 & 2 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }+2{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }+{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 4 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-\frac { 1 }{ 2 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 0 \end{matrix} \right] \)
Note
\(\left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 0 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow { R }_{ 2 }/8 }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} \right] \).
This is also a row-echelon form of the given matrix.
So, a row-echelon form of a matrix is not necessarily unique.

Given centre is (2, 3)
Let us solve 3x- 2y = 1  .....(1)
and 4x+ y = 27 .....(2)

⇒ x = 5
∴ 3(5) - 2y = 1
⇒ 15 - 2y = 1
⇒ 15-1 = 1
⇒ 14 = 2y
⇒ y = 7
 The circle passes through (5, 7)
[∵ distance between (5, 7) and (2, 3)]
r = \(\sqrt { { (5-2) }^{ 2 }+({ 7-3) }^{ 2 } } \)
= \(\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } \)
= \(\sqrt { 9+16 } =\sqrt { 25 } =5\)
Equation of the circle is
(x-h)²+(y-k)² = r²
(x - 2)² + (y - 3)² = 5²
x² - 4x + 4 + y² - 6y + 9 = 25
x² + y² - 4x - 6y + 13 - 25 = 0
x²+y²− 4x − 6y −12 = 0

Given adj A =\(\left[ \begin{matrix} 2 & -4 & 2 \\ -3 & 12 & -7 \\ -2 & 0 & 2 \end{matrix} \right] \)
We know that A = \(\pm \frac { 1 }{ \sqrt { |adjA| } } \) adj (adj A)..(1)
|adj A| = \(2\left| \begin{matrix} 12 & -7 \\ 0 & 2 \end{matrix} \right| +4\left| \begin{matrix} -3 & -7 \\ -2 & 2 \end{matrix} \right| +2\left| \begin{matrix} -3 & 12 \\ -2 & 0 \end{matrix} \right| \)
[Expanded along R₁]
= 2(24-0)+4(-6-14)+2(0+24)
= 2(24)+4(-20)+2(24) = 48-80+48
= 96-80 = 16
Now, adj (adj A)
=\(\left[ \begin{matrix} +\left| \begin{matrix} 12 & -7 \\ 0 & 2 \end{matrix} \right| & -\left| \begin{matrix} -3 & -7 \\ -2 & 2 \end{matrix} \right| & +\left| \begin{matrix} -3 & 12 \\ -2 & 0 \end{matrix} \right| \\ -\left| \begin{matrix} -4 & 2 \\ 0 & 2 \end{matrix} \right| & +\left| \begin{matrix} 2 & 2 \\ -2 & 2 \end{matrix} \right| & -\left| \begin{matrix} 2 & -4 \\ -2 & 0 \end{matrix} \right| \\ +\left| \begin{matrix} -4 & 2 \\ 12 & -7 \end{matrix} \right| & -\left| \begin{matrix} 2 & 2 \\ -3 & -7 \end{matrix} \right| & +\left| \begin{matrix} 2 & -4 \\ -3 & 12 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(24-0)-(6-14)+(0+24) \\ -(-8-0)+(4+4)-(0-8) \\ +(28-24)-(-14+6)+(24-12) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 24 & 20 & 24 \\ 8 & 8 & 8 \\ 4 & 8 & 12 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 24 & 8 & 4 \\ 20 & 8 & 8 \\ 24 & 8 & 12 \end{matrix} \right] \)
= \(4\left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)
Substituting (2) and (3) in (1) we get,
A = \(\frac { 1 }{ \sqrt { 16 } } .4\left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)
A = \(\pm \frac { 4 }{ 4 } \left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] =\pm \left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)

Equation of tangent to the circle at P(x₁, y₁ ) is xx₁ yy₁ = a²
That is, x(−3) + y(4) = 25
−3x + 4y = 25
Equation of normal is xy₁ - yx₁ = 0
That is, 4x + 3y = 0 

This an indeterminate of the form 1∞
Let \(g(x)={ \left( 1+\frac { 1 }{ x } \right) }^{ x }\)
Taking logarithm, we get,
\(log(g(x))=log{ \left( 1+\frac { 1 }{ x } \right) }^{ x }\)
\(=xlog{ \left( 1+\frac { 1 }{ x } \right) }\)
\(=\frac { { \left( 1+\frac { 1 }{ x } \right) } }{ \frac { 1 }{ x } } \)
\(\therefore ​​​​​​\underset { x\rightarrow \infty }{ lim } log(g(x))=​​​​​​\underset { x\rightarrow \infty }{ lim } \frac { { \left( 1+\frac { 1 }{ x } \right) } }{ \frac { 1 }{ x } } =\left( \frac { 0 }{ 0 } form \right) \)

\(=​​​​​​\underset { x\rightarrow \infty }{ lim } \frac { { 1 } }{ \left( 1+\frac { 1 }{ x } \right) } =1\)
But \(​​​​​​\underset { x\rightarrow \infty }{ lim } log(g(x))=log​​​​​​\underset { x\rightarrow \infty }{ lim } g(x)\)
\(\therefore log(\underset { x\rightarrow \infty }{ lim } g(x))=1\)
\({ e }^{ log }(\underset { x\rightarrow \infty }{ lim } g(x))={ e }^{ 1 }=e\Rightarrow (\underset { x\rightarrow \infty }{ lim } g(x))=e\)
\(\Rightarrow \underset { x\rightarrow \infty }{ lim } { \left( 1+\frac { 1 }{ x } \right) }^{ x }=e\)

f (x) is defined and differentiable for all x∈(-∞, ∞)
\(f'\left( x \right) =\frac { 1 }{ 2 } \left( { e }^{ x }+{ e }^{ -x } \right) \)
\(f''(x)=\frac { 1 }{ 2 } \left( { e }^{ x }-{ e }^{ -x } \right) \)
f"(x) = 0
\(\Rightarrow \frac { 1 }{ 2 } \left( { e }^{ x }-{ e }^{ -x } \right) =0\Rightarrow { e }^{ x }-{ e }^{ -x }=0\) 
\(\Rightarrow { e }^{ x }={ e }^{ -x }\Rightarrow { e }^{ x }=\frac { 1 }{ { e }^{ x } } \) 
\(\Rightarrow { e }^{ 2x }=1\Rightarrow { e }^{ 2x }={ e }^{ 0 }\) 
\(\Rightarrow 2x=0\Rightarrow x=0\) 
The possible intervals are (-∞,0) and (0,∞)

ஃ f(x) is concave up in (0, ∞) and concave down in (∞, 0).
Since f"(x) changes its position from negative to positive, when it passes through x = 0 the points of inflection is (0,1(0))
\(f(0)=\frac { 1 }{ 2 } \left( { e }^{ o }-{ e }^{ o } \right) =\frac { 1 }{ 2 } \left( 1-1 \right) =0\) 
ஃ (0, 0) is the point of inflection.

f(x) is defined and differentiable for all x ∈ (0, ∞).
\(\therefore f'(x)=\frac { { 3x }^{ 2 } }{ 3 } -\frac { 1 }{ x } ={ x }^{ 2 }-\frac { 1 }{ x } \) 
\(\Rightarrow { x }^{ 2 }-\frac { 1 }{ x } =0\Rightarrow \frac { { x }^{ 3 }-1 }{ x } =0\)
\(\Rightarrow { x }^{ 3 }-1=0\Rightarrow { x }^{ 3 }1\) 
\(\Rightarrow\) x = 1
\(\therefore\) The Stationary point is at x = 1
The possible intervals are (0, 1) and (1, ∞).

Since f '(x) changes from negative to positive at x = 1, there is a local minimum at x = 1.
\(\therefore f'\left( 1 \right) =\frac { { 1 }^{ 3 } }{ 3 } -log1\) 
= \(\frac { 1 }{ 3 } -o=\frac { 1 }{ 3 } \)

Given f(x) = 2.0 + 3x2 - 12x
The given function is defined and differentiable for all x ∈(-∞,∞)
f'(x) = 6x² + 6x - 12
The stationary points are given by
6x² + 6x - 12 = 0
\(\Rightarrow\) x² +x-2 = 0
\(\Rightarrow\) (x + 2)(x - 1) = 0
\(\Rightarrow\) x = -2, 1
Hence the intervals of monotonicity are
(-∞, - 2), (-2, 1) and (1, ∞).

ஃ f(x) is strictly increasing on (-∞, - 2)
(1, ∞) and strictly decreasing on (-2, 1).
Since f'(x) changes from positive to negative at x = -2, there is a local maximum at x = -2
\(\therefore\) f(-2) = 2 (2)³ + 3 (2)² - 12 (-2) 
= 2(-8) + 3(4) + 24 
= -16+12+24 = 20
Also f'(x) changes from negative to positive at x = 1, there is a local minimum at x = 1.
ஃ f(1) = 2 (1)³ + 3 (1)² - 12(1)
= 5 -12 = -7

9x²- y²- 36x - 6y + 18 = 0
Given equation is 9x²- y²- 36x - 6y + 18 = 0
⇒ 9x² - 36x - (y² + 6y) = -18
⇒ 9(x²-4x)-(y²+6y) =-18
⇒ 9(x² - 4x + 4 - 4) - (y² + 6y + 9 - 9) = -18
⇒ 9(x-2)²-36-(y+3)²+9 =-18
⇒ 9(x-2)² - (y+3)² = -18+36-9
⇒ 9(x - 2)² - (y + 3)² = 9
Dividing by 9 we get, \(\frac { { (x-2) }^{ 2 } }{ 1 } -\frac { ({ y+3) }^{ 2 } }{ 9 } =1\)
This is an equation of the hyperbola whose transverse axis is parallel to x-axis.
a² = 1, b² = 9
∴ c² = a² + b² = 1 + 9 = 10 ⇒ c = \(\sqrt { 10 } \)
\(e=\sqrt {1-\frac { { b }^{ 2 } }{ { a }^{ 2 } }} =\sqrt { 1-\frac { 9 }{ 1 } } =\sqrt { 10 } \)
a) Center is (2, -3)
⇒ h = 2, k = -3
(b) Foci are (h + c, k), (17 - c, k)
⇒ (2 +\(\sqrt { 10 } \), -3), (2 - \(\sqrt { 10 } \), -3)
(c) Vertic ar (h + a, k) (h - a, k)
⇒ (2+ 1,-3), (2-1,-3)
⇒ (3, -3) (1, -3)
(d) Equation of directrices are x - 2 = \(\pm \frac { a }{ e } \)
⇒ \(x-2=\pm \frac { 1 }{ \sqrt { 10 } } \)
\(x=2\pm \frac { 1 }{ \sqrt { 10 } } \)
⇒ \(x=2+\frac { 1 }{ \sqrt { 10 } } \) and \(x=2-\frac { 1 }{ \sqrt { 10 } } \)

Given equation is \(\frac { { (x-1) }^{ 2 } }{ 100 } +\frac { ({ y-2) }^{ 2 } }{ 64 } =1\)
This is an equation of the ellipse
∴ a² = 100,b² = 64
⇒ c² = a² - b²
⇒ c² = 100 - 64 = 36
∴ c = 6
\(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 64 }{ 100 } } =\sqrt { \frac { 100-64 }{ 100 } } \)
\(=\sqrt{\frac{36}{100}}=\frac{\not 6} {\not {10}}\)
∴ c = \(\frac { 3 }{ 5 } \)
(a) Center is (-1, 2) ⇒ h = -1, k = 2
(b) Foci are (h - c, k), (h + c, k) ⇒ (-1 - 6, 2), (-1 + 6, -2) ⇒ (-7, 2), (5, 2)
(c) Vertices are (h - a, k) and (h + a, k) ⇒ (-1 -10,2), (-1 + 10,2) ⇒ (-11, 2), (9, 2)
(d) Equation of directrices are x + 1 = \(\pm \frac { a }{ e } \)
\(\Rightarrow x+1=\pm \frac { 10 }{ \frac { 3 }{ 5 } } \Rightarrow x+1=\pm \frac { 50 }{ 3 } \)
\(\therefore x+1=\frac { 50 }{ 3 } and x+1=-\frac { 50 }{ 3 } \)
\(\Rightarrow x=\frac { 50 }{ 3 } -1\) and \(x=-\frac { 50 }{ 3 } -1\)
\(\Rightarrow x=\frac { 47 }{ 3 } \) and \(x=\frac { -53 }{ 3 } \)

Let the equation of the ellipse be
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Length of semi minor axis b = 5
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Let BB' be the road width and AA' be the end points of the opening of the tunnel.
Let CB = 8, BD = 1,
D is (8, 4) lies on the ellipse
\( \frac{8^{2}}{a^{2}}+\frac{4^{2}}{5^{2}}=1 \)
\( \Rightarrow a^{2}=\frac{25}{9} \times 64 \)
\( \Rightarrow a=\frac{40}{3} \)
The width AA' = 2a
\(=\frac{80}{3}=26.66 \mathrm{~m}\)
The required width is 26.66 m.

2x - y + z = 2, 6x - 3y + 3z = 6, 4x - 2y + 2z = 4
The matrix form of the given system is AX = B where
A =\(\left[ \begin{matrix} 2 & -1 & 1 \\ 6 & -3 & 3 \\ 4 & -2 & 2 \end{matrix} \right] \)
\(X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] B=\left[ \begin{matrix} 2 \\ 6 \\ 4 \end{matrix} \right] \)
Applying elementary row operations on the augment matrix [A|B], we get,
[A|B] =\(\left[ \begin{matrix} 2 & -1 & 1 \\ 6 & -3 & 3 \\ 4 & -2 & 2 \end{matrix}|\begin{matrix} 2 \\ 6 \\ 4 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 2 & -1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 2 \\ 0 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 1 [∵ only one non zero row]
and \(\rho \)[A|B] = 1 [∵ only one-zero row]
∴ \(\rho \)(A) =\(\rho \)(A|B] = 1< 3 the given system is consistent and has two parameter family of solutions.
So, z = t and y = s where, t \(\in \)R
Writing the equivalent equations from the rowechelon matrix, we get
2x-y+z = 2 .............(1)
y = s
z = t
Substituting (2) and (3) In (1) we get
2x-s+t = 2
⇒ 2x-s+t = 2
⇒ x = \(\frac{1}{2}\)[s-t+2]
∴ Solution set is {\(\frac{1}{2}\)(s-t+2),s,t} here s, t \(\in \) R.

Here the number of unknowns is 3.
The matrix form of the system is AX = B, where
A = \(\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & -3 & 3 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} -9 \\ -18 \\ -27 \end{matrix} \right] \).
Applying elementary row operations on the augmented matrix[A | B], we get
[A | B] = \(\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & -3 & 3 \end{matrix}|\begin{matrix} -9 \\ -18 \\ -27 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-2{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} -9 \\ 0 \\ 0 \end{matrix} \right] \).
So, ρ(A) = ρ ([A | B]) = 1 < 3.
From the echelon form, we get the equivalent equations x - y + z = -9, 0 = 0, 0 = 0.
The equivalent system has one non-trivial equation and three unknowns.
Taking y = s, z = t arbitrarily, we get x - s + t = -9; x = -9 + s - t.
So, the solution is (x = -9 + s - t, y = s, z = t), where s and t are parameters.
The above solution set is a two-parameter family of solutions.
Here, the given system of equations is consistent and has infinitely many solutions which form a two parameter family of solutions.

x²+3y² = 12
\(\div 12\) we get, \(\frac { { x }^{ 2 } }{ 12 } +\frac { { y }^{ 2 } }{ 4 } =1\)
∴ a² = 12, b² = 4
The line x-y+ 4 = 0 can be rewritten as y = x+4.
∴ m = 1, c = 4
The condition for y = mx + 4 to be a tangent to the ellipse is c² = a²m² + b²
∴ (4)² = 12(1)² + 4
⇒ 16 = 12+4
⇒ 16 = 16
Since the condition is satisfied, the line x - y + 4 = 0 is a tangent to the ellipse x² + 3y² = 12.
Also, the point of contact is \(\left( -\frac { { a }^{ 2 }m }{ c } ,\frac { { b }^{ 2 } }{ c } \right) \)
\(\Rightarrow \left( -\frac { 12(1) }{ 4 } ,\frac { 4 }{ 4 } \right) \Rightarrow (-3,1)\)
∴The point of contact is (-3, 1).

2x − 2y + 3z = 2, x + 2y − z = 3, 3x − y + 2z = 1
Transforming the augmented matrix to echelon form, we get
\(\left[ \begin{matrix} 2 & -2 & 3 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{matrix}|\begin{matrix} 2 \\ 3 \\ 1 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \)
\(\left[ \begin{matrix} 1 & 2 & -1 \\ 2 & -2 & 3 \\ 3 & -1 & 2 \end{matrix}|\begin{matrix} 3 \\ 2 \\ 1 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & -1 \\ 0 & -6 & 5 \\ 0 & -7 & 5 \end{matrix}|\begin{matrix} 3 \\ -4 \\ -8 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & -1 \\ 0 & -6 & 5 \\ 0 & -1 & 0 \end{matrix}|\begin{matrix} 3 \\ -4 \\ -4 \end{matrix} \right] \overset { { R }_{ 3 }\rightarrow { 6R }_{ 3 }-2{ R }_{ 2 } }{ \longrightarrow } \)
\(\left[ \begin{matrix} 1 & 2 & -1 \\ 0 & -6 & 5 \\ 0 & 0 & -5 \end{matrix}|\begin{matrix} 3 \\ -4 \\ -20 \end{matrix} \right] \)
Writing the equivalent equations from the rowechelon matrix, we get,
x+2y-z = 3 .......... (1)
-6y + 5z =-4 ..........(2)
-5z = -20 ⇒ z =\(\frac{-20}{-5}\) = 4
Substituting z = 4 in 2 we ge
-6y + 5(4) = -4
⇒ -6y+20 = -4 ⇒ -4 - 20 = -20
⇒ y = \(\frac{-24}{-6}\) = 4
Substituting y = z = 4 in (1) we get
x+ 2(4) -4 = 3
⇒ x + 8 - 4 = 3
⇒ x + 4 = 3
⇒ x = 3 - 4 = -1
∴ Solution set is {-1, 4, 4}

Let the cost of one dosa be Rs. x
The cost of one idli be Rs. y
and the cost of one vadai be Rs. z
By the given data,
2x+ 3y + 2z = 150
2x + 2y + 4z = 200
5x + 4y + 2z = 250
∴ Δ = \(\left| \begin{matrix} 2 & 3 & 2 \\ 2 & 2 & 4 \\ 5 & 4 & 2 \end{matrix} \right| \)
= \(2\left| \begin{matrix} 2 & 4 \\ 4 & 2 \end{matrix} \right| -3\left| \begin{matrix} 2 & 4 \\ 5 & 2 \end{matrix} \right| +2\left| \begin{matrix} 2 & 2 \\ 5 & 4 \end{matrix} \right| \)
= 2(4 - 16) - 3(4 - 20) + 2(8 - 10)
= 2(- 12) - 3(- 16) + 2(- 2)
= - 24 + 48 - 4 = 20
Δ₁ = \(\left| \begin{matrix} 150 & 3 & 2 \\ 200 & 2 & 4 \\ 250 & 4 & 2 \end{matrix} \right| \)
Taking 50 common from C3 we get,
= 100\(\left| \begin{matrix} 3 & 3 & 1 \\ 4 & 2 & 2 \\ 5 & 4 & 1 \end{matrix} \right| \)
= \(100\left[ 3\left| \begin{matrix} 2 & 2 \\ 4 & 1 \end{matrix} \right| -3\left| \begin{matrix} 4 & 2 \\ 5 & 1 \end{matrix} \right| +1\left| \begin{matrix} 4 & 2 \\ 5 & 4 \end{matrix} \right| \right] \)
= 100[3(2 - 8) - 3(4 - 10) + 1(16 - 10)]
= 100[3(-6) - 3(- 6) + 6]
= 100[- 18 + 18 + 6] = 600
Δ₂ = \(\left| \begin{matrix} 2 & 150 & 2 \\ 2 & 200 & 4 \\ 5 & 250 & 2 \end{matrix} \right| =100\left| \begin{matrix} 2 & 3 & 1 \\ 2 & 4 & 2 \\ 5 & 5 & 1 \end{matrix} \right| \)
= \(100\left[ 3\left| \begin{matrix} 4 & 2 \\ 5 & 1 \end{matrix} \right| -3\left| \begin{matrix} 2 & 2 \\ 5 & 1 \end{matrix} \right| +1\left| \begin{matrix} 2 & 4 \\ 5 & 5 \end{matrix} \right| \right] \)
= 100[2(4 - 10) - 3(2 - 10) + 1(10 - 20)]
= 100[2(- 6) - 3(- 8) + 1(- 10)]
= 100[- 12 + 24 - 10] = 100 [2] = 200
Δ₃ = \(\left| \begin{matrix} 2 & 3 & 150 \\ 2 & 2 & 200 \\ 5 & 4 & 250 \end{matrix} \right| =50\left| \begin{matrix} 2 & 3 & 3 \\ 2 & 2 & 4 \\ 5 & 4 & 5 \end{matrix} \right| \)
= \(50\left[ 2\left| \begin{matrix} 2 & 4 \\ 4 & 5 \end{matrix} \right| -3\left| \begin{matrix} 2 & 4 \\ 5 & 5 \end{matrix} \right| +3\left| \begin{matrix} 2 & 2 \\ 4 & 4 \end{matrix} \right| \right] \)
= 50 [2(10 - 16) - 3(10 - 20) + 3(8 - 10)]
= 50[2(- 6) - 3(- 10) +3(- 2)]
= 50 [- 12 + 30 - 6] = 50 [12] = 600
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 600 }{ 20 } \) = 30
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 200 }{ 20 } \) = 10
z = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 600 }{ 20 } \) = 30
Hence, the price of one dosa be Rs. 30, one idli be Rs. 10 and the price of 1 vadai be Rs. 30.
Also the cost on dosa, six idlies and six vadai is
= 3x + 6y + 6z = 3(30) + 6(10) + 6(30)
= 90 + 60 + 180 = Rs. 330
Since the family had Rs. 350 in hand, they will be able to manage to pay the bill.

\(\frac { { (x-3) }^{ 2 } }{ 225 } +\frac { ({ y-4) }^{ 2 } }{ 289 } =1\)
Given equation is \(\frac { { (x-3) }^{ 2 } }{ 225 } +\frac { ({ y-4) }^{ 2 } }{ 289 } =1\)
This is an equation of the ellipse a² = 289
b² = 225 and
c² = a² - b² ⇒ 289 - 225 = 64 ⇒ c = 8.
\(e=\sqrt { \frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 225 }{ 289 } } =\sqrt { \frac { 289-225 }{ 289 } } \)
 \(=\sqrt { \frac { 64 }{ 289 } } =\frac { 8 }{ 17 } \)
(a) Center is (3, 4) ⇒ h = 3, k = 4
(b) foci are (h, k+c), (h, k-c)
⇒ (3, 4 + 8), (3, 4 - 8) ⇒ (3, 12), (3,-4)
(c) Vertices are (h, k - a), (h, k + a)
⇒ (3, 4 -17), (3, 4 + 17) ⇒ (3, -13), (3, 21)
(d) Equations of directrices are y - 4 = \(\pm \frac { a }{ e } \)
\(\Rightarrow y-4=\pm \frac { 17 }{ \frac { 8 }{ 17 } } +4\Rightarrow y-4=\pm \frac { -289 }{ 8 } +4\)
\(\Rightarrow y=\frac { 289 }{ 8 } +4\) and \(y=\frac { -289 }{ 8 } +4\)
\(\Rightarrow y=\frac { 289+32 }{ 2 } \) and \(y=\frac { -289+32 }{ 8 } \)
\(\Rightarrow y=\frac { 321 }{ 8 } \) and \(y=\frac { -257 }{ 8 } \)

Rearranging terms in the equation of hyperbola to bring it to standard form,
we have, 11(x²-4x)-25(y²-2y)-256 = 0
11(x− 2)²−25(y−1)² = 256−44+25
11(x−2 )²− 25 (y−1)² = 275
\(\frac { { \left( x-2 \right) }^{ 2 } }{ 25 } -\frac { { \left( y-1 \right) }^{ 2 } }{ 11 } =1\)
Centre (2, 1) a² = 25, b² = 11
c² = a² +b²
= 25 +11 = 36
Therefore, c = ±6
and e = \(\frac { c }{ a } =\frac { 6 }{ 5 } \)and the coordinates of foci are(8, 1) and(-4, 1) from figure.