order 2 ; degree 1

Here, the highest order derivative is \(\frac { { d }^{ 4 }y }{ { dx }^{ 4 } } \) whose power is 3.
Therefore, the given differential equation is of order 4 and degree 3.

\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =xy+cos\left( \frac { dy }{ dx } \right) \)
The given differential equation is
\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =xy+cos\left( \frac { dy }{ dx } \right) \)
The highest derivative is 2
∴ Order 2.
The given differential equation is not a polynomial equations in its derivatives and so its degree is not defined.

\({ \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }-3\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { dy }{ dx } +4=0\)
Given differential equation is
\({ \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }-3\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) +5\frac { dy }{ dx } +4=0\)
\(\Rightarrow { \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }=3\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) -5\left( \frac { dy }{ dx } \right) -4\)
Taking power 3 both sides,
\(\Rightarrow { \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ 2 }={ \left( 3\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) -5\left( \frac { dy }{ dx } \right) -4 \right) }^{ 3 }\)
The highest derivative is 3 and its power is 2.
∴ Order 3, degree 2.

Let P(x, y) be the movable point.
By focal property of ellipse, PA + PB = 2a
∴ 2a = 10 ⇒ a = 5
Since focus is (4, 0), ae = 4 ⇒ 5e = 4 ⇒ e = \(\frac45\)
Also b² = a²(1 - e²) = 25\(\left( 1-\frac { 16 }{ 25 } \right) =25\left( \frac { 9 }{ 25 } \right) \) = 9
Equation of ellipse is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\)

Let p(x) = x⁷-6x⁶+7x⁵+5x²+2x+2
It has only one change of sign. Now,
p(-x) = (-x)⁷ -6(-x)⁶ +7(-x)⁵ +5(-x)² +2(-x)+2
= -x⁷ -6x⁶ -7x⁵ + 5x² - 2x + 2
It has two, change of sign.
Hence, p(x) has one positive root and has at least two negative roots.

Here A = 0, B = 0, C = 1, D = 4, E = 3, F = 4
B = 0, A = 0 either A or C is 0.
Hence, the given equation represents a parabola.

Here Δ = b²−4ac = 0 for equal roots. This implies 4(k + 2)² = 4(9)k. This implies k = 4 or 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 α and β are the roots of 17x²+ 43x −73 = 0 , we have α + β =\(\frac { -43 }{ 17 } \) and  αβ =\(\frac { -73 }{ 17 } \).
We wish to construct a quadratic equation with roots α + 2 and  β + 2.Thus, to construct such a quadratic equation, calculate,
the sum of the roots =  α + β + 4 = \(\frac { -4 }{ 17 } +4=\frac { 25 }{ 17 } \) and
the product of the roots = αβ + 2(α+β)+4 = \(\frac { -73 }{ 17 } +2\left( \frac { -43 }{ 17 } \right) +4=\frac { -91 }{ 17 } \)
Hence a quadratic equation with required roots is x²-\(\frac { 25 }{ 17 } x-\frac { 91 }{ 17 } \) = 0
Multiplying this equation by 17, gives 17x²−25x−91 = 0
which is also a quadratic equation having roots α + 2 and β + 2

\( \frac{d v}{d t} =g \cos \alpha-k v \)
\(\frac{d v}{d t}+\mathrm{k} v =g \cos \alpha \)
Which is of the form
\( \frac{d v}{d t}+\mathrm{p} v =\mathrm{Q} \)
Where \(\mathrm{P} =\mathrm{k}_{1} \mathrm{Q}=\mathrm{g} \cos \alpha \)
\(\mathrm{I} \cdot \mathrm{F} =\mathrm{e}^{\mathrm{P} d t} \)
\( =\mathrm{e}^{\int \mathrm{kdt}}\)
The general solution
\( \mathrm{y}(\mathrm{I} . \mathrm{F}) =\int \mathrm{Q}(\mathrm{I} . \mathrm{F}) \mathrm{dx}+\mathrm{c} \)
\(\mathrm{v}(\mathrm{I} . \mathrm{F}) =\int \mathrm{Q}(\mathrm{I} . \mathrm{F}) \mathrm{dt}+\mathrm{c} \)
\(\mathrm{Ve}^{\mathrm{kt}} =\int \mathrm{g} \cos \alpha \mathrm{e}^{\mathrm{kt}} \mathrm{dt}+\mathrm{c} \)
\(\mathrm{Ve}^{\mathrm{kt}} =\int \mathrm{g} \cos \alpha\left(\frac{\mathrm{e}^{\mathrm{kt}}}{\mathrm{k}}\right)+\mathrm{c} \)
put v = 0, t = 0
\(0 =g \cos \alpha\left(\frac{\mathrm{e}^{0}}{\mathrm{k}}\right)+\mathrm{c} \)
\(\mathrm{c} =-\frac{\mathrm{g} \cos \alpha}{\mathrm{k}} \)
\(\mathrm{Ve}^{\mathrm{kt}} =\mathrm{g} \cos \alpha\left(\frac{\mathrm{e}^{\mathrm{kt}}}{\mathrm{k}}\right)-\frac{g \cos \alpha}{\mathrm{k}} \)
\(\mathrm{Ve}^{\mathrm{kt}} =\frac{\mathrm{g} \cos \alpha}{\mathrm{k}}\left(\mathrm{e}^{\mathrm{kt}}-1\right) \)
\(\mathrm{V} =\frac{\mathrm{g} \cos \alpha}{\mathrm{k}}\left(1-\mathrm{e}^{-\mathrm{kt}}\right) \)

Rearranging terms in the equation of hyperbola to bring it to standard form,
We have,
\( 12\left(x^{2}-2 x\right)-4\left(y^{2}-8 y\right)-124 =0 \)
\(12(x-1)^{2}-4(y-4)^{2} =124+12-64 \)
\(12(x-1)^{2}-4(y-4)^{2} =72 \)
\(\frac{(x-1)^{2}}{6}-\frac{(y-4)^{2}}{18} =1\)
Centre (1, 4); a² = 6;
\( b^{2} =18 \)
\(c^{2} =a^{2}+b^{2} \)
\( =6+18 \)
\(c^{2} =24 \)
\(c =\pm 2 \sqrt{6} \)
\(e =\frac{c}{a}=\frac{2 \sqrt{6}}{6}\)
eccentricity, \(e=\frac{\sqrt{6}}{3}\)
centre (h, k) = (1, 4)
foci \( =(h, \pm c+k) \)
\(=(1, \pm 2 \sqrt{6}+4) \)
\( =(1,2 \sqrt{6}+4) 8(1,-2 \sqrt{6}+4)\)

Let T be the temperature of the body at any time t and with time 0 taken to be 8 p.m. By Newton’s law of cooling \(\frac { dT }{ dt } =k(T-50)or\frac { dT }{ T-50 } =dt\).
Integrating on both sides, we get log |50 −T| = kt + logC or 50 −T = Ce^kt.
When t = 0, T = 70, and so C = −20
When t = 2,T = 60, we have −10 = −20 e^k2.
Thus, \(k=\frac { 1 }{ 2 } log\left( \frac { 1 }{ 2 } \right) \)
Hence, the solution is 50-T = -20e^{\(\frac{1}{2}\)tlog\((\frac{1}{2})\)} or T = 50 + 20\((\frac{1}{2})^\frac{t}{2}\)
Now, we would like to find the value of t, for which T(t) = 98.6 , and t = 2\(\left( \frac { log\left( \frac { 48.6 }{ 20 } \right) }{ log\left( \frac { 1 }{ 2 } \right) } \right) \approx -2.56\)
It appears that the person was murdered at about 5.30 p.m.

\(\frac { dy }{ dx } +\frac { { 3x }^{ 2 }y }{ 1+{ x }^{ 3 } } =\frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } \)
This is a linear differential equation 
\(\therefore P=\frac { { 3x }^{ 2 } }{ 1+{ x }^{ 3 } } ;Q=\frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } \)
\(\therefore \int { pdx } =\int { \frac { { 3x }^{ 2 } }{ 1+{ x }^{ 3 } } dx } =log(1+{ x }^{ 3 })\)
\(\therefore I.F.={ e }^{ \int { pdx } }={ e }^{ log(1+{ x }^{ 3 }) }=(1+{ x }^{ 3 })\)
\(\therefore\)The solution is \({ ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow y(1+{ x }^{ 3 })=\int { \frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } (1+{ x }^{ 3 })dx+c } \)
\(cos2x=1-2{ sin }^{ 2 }x\)
\(sin2x=\frac { 1-cos2x }{ 2 } =\int { { sin }^{ 2 }xdx+c } \)
\(\Rightarrow y(1+{ x }^{ 3 })=\int { \frac { 1-cos2x }{ 2 } } dx+c\)
\(\Rightarrow y(1+{ x }^{ 3 })=\frac { x }{ 2 } -\frac { sin2x }{ 4 } +c\)

The given equation can be written as \(\frac { dy }{ dx } =\frac { 2x+3y }{ x-y } \)
This is a homogeneous equation.
Let y = vx . Then we have \(v+x\frac { dv }{ dx } =\frac { 2+3v }{ 1-v } \)
Thus, \(x\frac { dv }{ dx } =\frac { 2+2v+{ v }^{ 2 } }{ 1-v } or\frac { 1-v }{ { (1+v) }^{ 2 }+1 } dv=\frac { dx }{ x } or\frac { 1 }{ 2 } \left[ \frac { 2v+2 }{ { v }^{ 2 }+2v+2 } -\frac { 4 }{ { (v+1) }^{ 2 }+1 } \right] dv=\frac { dx }{ x } \)
Integrating both sides, we get -\(\frac{1}{2}\)log |v²+2v+2|+2tan^-1(v+1) = log |x| + log |C|
or log |v²+2v+2| -4tan^-1(v+1) = -2log |x| -2 log |C|
or log |v²+2v+2| +log |x|² - 4tan^-1(v+1) = -2 log |C|
or log |(v²+2v+2)x²| -4 tan^-1(v+1) = -2 log |C|
Now replacing v by\(\frac{y}{x},\) we get, log |y²+2xy+2x²|-4tan^-1\(\left( \frac { x+y }{ x } \right) =k\), where k = -2 log |C| gives the required solution.

\(\Rightarrow \frac { y\quad dx-x\quad dy }{ { y }^{ 2 } } .cot\left( \frac { x }{ y } \right) =xdx\)
put \(\frac { x }{ y } =t\)
\(\Rightarrow \frac { y\quad dx-x\quad dy }{ { y }^{ 2 } } =dt\)
Substituting these values in equation (1), we get
dt cot(t) = x dx
 cot t dt = ndx
Taking integration on both sides, we get
\(\Rightarrow \int { cot(t)dt=n\int { dx } } \)
\( \int \cot t \mathrm{dt} =n \int d x \)
\(\log (\sin \mathrm{t}) =\mathrm{n} x+\mathrm{C}_1 \)
\(\sin \mathrm{t} =\mathrm{e}^{\mathrm{nx}+\mathrm{c}_1} \)
\(\sin \left(\frac{x}{y}\right) =\mathrm{e}^{n x} \mathrm{e}^{\mathrm{C}_r} \)
\(\sin \left(\frac{x}{y}\right) =\mathrm{C}^{\mathrm{nx}}\)
\(\\ \Rightarrow sin\left( \frac { x }{ y } \right) ={ e }^{ nx+c }\left[ \because t=\frac { x }{ y } \right] \)

\(\frac { { (y-2) }^{ 2 } }{ 25 } -\frac { ({ x+1) }^{ 2 } }{ 16 } =1\)
Given equatlon is \(\frac { { (y-2) }^{ 2 } }{ 25 } -\frac { ({ x+1) }^{ 2 } }{ 16 } =1\)
Thi is an equation of the hyperbola where transverse axis is parallel to y-axis.
∴ a² = 25, b² = 16
⇒ c² - a² + b² = 25 +16 = 41
⇒ c = \(\sqrt { 41 } \)
\(e =\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1+\frac { 16 }{ 25 } } =\sqrt { \frac { 41 }{ 25 } } =\frac { \sqrt { 41 } }{ 5 } \)
(a) Center is (-1, 2) ⇒ h = -1, k = 2
(b) Foci are (h, k + c), (h, k- c)
⇒ (-1, 2 + \(\sqrt { 41 } \)), (-1, 2 - \(\sqrt { 41 } \))
(c) Vertices are (h, k + a), (h, k-a)
⇒ (-1, 2 + 5), (-1, 2 - 5)
= (-1, 7), (-1, -3)
(d) Equation of directrices are
\(y-2=\pm \frac { 5 }{ \frac { \sqrt { 41 } }{ 5 } } \)
\(\Rightarrow y-2=\pm \frac { 25 }{ \sqrt { 41 } } \)
\(\Rightarrow y=2+\frac { 25 }{ \sqrt { 41 } } \) and 
\( y=2-\frac { 25 }{ \sqrt { 41 } } \)

The cross section of a nuclear cooling tower is in the shape of a hyperbola.

GIven OC = \(\frac12\) OD and CD = 150 m
Its equation is OC = 50 m & OD = 100 m
\(\frac { { x }^{ 2 } }{ { 30 }^{ 2 } } -\frac { { y }^{ 2 } }{ { 44 }^{ 2 } } =1\)  ......(1)
Let I be the radius of the top of the tower
∴ A(l, 50) is a point on the hyperbola
∴ \(\frac { { l }^{ 2 } }{ { 30 }^{ 2 } } \frac { { 50 }^{ 2 } }{ { 44 }^{ 2 } } =1\)
⇒ \(\frac { { l }^{ 2 } }{ { 30 }^{ 2 } } =1+\frac { { 50 }^{ 2 } }{ { 44 }^{ 2 } } =\frac { { 44 }^{ 2 }+{ 50 }^{ 2 } }{ { 44 }^{ 2 } } \)
⇒ \({ l }^{ 2 }=\frac { { 30 }^{ 2 } }{ { 44 }^{ 2 } } \) (1936 + 2500)
⇒ l² = \(\frac { 30 }{ 40 } \sqrt { 4436 } =\frac { 30 }{ 44 } \)(66.60)
⇒ \(\frac { 30 }{ 40 } \sqrt { 4436 } =\frac { 30 }{ 44 } \) = 45.41 m
Radius of the top of the tower is 45.41 m.
Let h be the radius of the base of the tower.
∴ B(h, 100) is a point on the hyperbola
∴ (1) becomes
\(\frac { { h }^{ 2 } }{ { 30 }^{ 2 } } -\frac { { 100 }^{ 2 } }{ { 44 }^{ 2 } } =1\Rightarrow \frac { h^{ 2 } }{ { 30 }^{ 2 } } =1+\frac { { 100 }^{ 2 } }{ { 44 }^{ 2 } } =\frac { { 44 }^{ 2 }+{ 100 }^{ 2 } }{ { 44 }^{ 2 } } \)
⇒ \({ h }^{ 2 }=\frac { { 30 }^{ 2 } }{ { 44 }^{ 2 } } \) (1936 + 10000)
⇒ \({ h }^{ 2 }=\frac { 30 }{ 44 } \sqrt { 11936 } =\frac { 30 }{ 44 } \)(109.25)
⇒ \(h=\frac { 3277.5 }{ 44 } \) = 74.48 m.
Radius of the base of the tower is 74.48 m.
Diameter of the base = 148.96 m
Diameter of the top and base of the tower are 90.82 m and 148.96 m.

Equation of the hyperbola is 12x²- 9y² = 108
\(\div 108\) we get, \(\frac { { 12x }^{ 2 } }{ 108 } -\frac { 9{ y }^{ 2 } }{ 108 } =1\)
⇒ \(\frac { { x }^{ 2 } }{ 9 } -\frac { { y }^{ 2 } }{ 12 } =1\)
∴ a² = 9, b² = 12
Parametric equation of tangent to the hyperbola is \(\frac { x \ sec\ \theta }{ a } -\frac { y \ tan \ \theta }{ b } =1\)
When \(\theta =\frac { \pi }{ 3 } \), the equation is
\(\frac { xsec\frac { \pi }{ 3 } }{ 3 } -\frac { ytan\frac { \pi }{ 3 } }{ 2\sqrt { 3 } } =1\)

⇒ \(\frac { 4x-3y }{ 6 } =1\) ⇒ 4x - 3y - 6 = 0 is the required equation of tangent.
Parametric equation of normal to the hyperbola is
\(\frac { ax }{ sec\theta } +\frac { by }{ tan\theta } ={ a }^{ 2 }+{ b }^{ 2 }\)
At \(\theta =\frac { \pi }{ 3 } ,\frac { 3x }{ sec\frac { \pi }{ 3 } } +\frac { 2\sqrt { 3 } }{ tan\frac { \pi }{ 3 } } =9+12\)
\(\Rightarrow \frac{3 x}{2}+\frac{2 \sqrt{\not 3} y}{\sqrt{\not 3}}=21\)
⇒ \(\frac { 3x }{ 2 } \) + 2y = 21 ⇒ 3x + 4y = 42
⇒ 3x + 4y - 42 = 0 is the required equation of normal.

\(8x^{ \frac { 3 }{ 2x } }-8x^{ \frac { -3 }{ 2x } }\) = 63
\(\Rightarrow 8\left[ { \left( { x }^{ \frac { 1 }{ 2n } } \right) }^{ 3 }-{ \left( { x }^{ \frac { -1 }{ 2n } } \right) }^{ 3 } \right] =63\)
Put \({ x }^{ \frac { 1 }{ 2n } }=y\)
\(\Rightarrow 8\left( { y }^{ 2 }-\frac { 1 }{ { y }^{ 3 } } \right) =63\)
\(\Rightarrow { y }^{ 3 }-\frac { 1 }{ { y }^{ 3 } } =\frac { 63 }{ 8 } \Rightarrow \frac { { y }^{ 6 }-1 }{ { y }^{ 3 } } =\frac { 63 }{ 8 } \)
\(\Rightarrow { 8y }^{ 6 }-8=63{ y }^{ 3 }\)
\(\Rightarrow { 8y }^{ 6 }-{ 63y }^{ 3 }-8=0\)
\(\Rightarrow { 8t }^{ 2 }-63t-8=0\ [where\quad t={ y }^{ 3 }]\)
\(\Rightarrow (8t-1)(t-8)=0\)
\(\Rightarrow t=\frac { 1 }{ 8 } ,8\)
Case (i): when \(t=8,\Rightarrow { y }^{ 3 }=8\Rightarrow { y }^{ 2 }={ 2 }^{ 3 }\)
\(\Rightarrow y=2\)
Case (ii): when \(t=\frac { 1 }{ 8 } ,{ y }^{ 3 }=\frac { 1 }{ 8 } \Rightarrow y=\frac { 1 }{ 2 } \)
When \(y=2,{ x }^{ \frac { 1 }{ 2n } }=2\)
\(\Rightarrow x={ (2 })^{ 2n }\quad \Rightarrow x={ ({ 2 }^{ 2 }) }^{ n }\)
\(\Rightarrow x={ 4 }^{ n }\)
When \(y=\frac { 1 }{ 2 } ,{ x }^{ \frac { 1 }{ 2n } }=\frac { 1 }{ 2 } \Rightarrow x={ \left( \frac { 1 }{ 2 } \right) }^{ 2n }\)
\(\Rightarrow x={ \left( \frac { 1 }{ { 2 }^{ 2 } } \right) }^{ n }=\frac { 1 }{ { 4 }^{ n } } \)
Hence the roots are 4ⁿ.

This equation is Type I even degree reciprocal equation. Hence it can be rewritten as
x²\(\left[ \left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) -10 \left( x+\frac { 1 }{ x } \right) +26 \right] \)= 0 Since x \(\neq\) 0, we get \(\left(x^{2}+\frac{1}{x^{2}}\right)-10\left(x+\frac{1}{x}\right)+26=0\)
Let y = \(\left( x+\frac { 1 }{ x } \right) \)
[(y²-2)-10y+26] = 0 ⇒ (y²-10y+24) = 0 ⇒ (y-6)(y-4) = 0 ⇒ y = 6 or y = 4
Case (i)
y = 6 ⇒ x +\(\frac{1}{x}\) = 6 ⇒ x = 3+2\(\sqrt{2}\), x = 3 - 2\(\sqrt{2}\)
Case (ii)
y = 4 ⇒ x = 2+\(\sqrt{3}\), x = 2-\(\sqrt{3}\).
Hence, the roots are \(3 \pm 2 \sqrt{2}, 2 \pm \sqrt{3}\)

Let the general equation of the circle be
x² +y² +2gx + 2fy + c = 0 ........ (1)
It passes through points (1, 1), (2, -1) and(3, 2) .
Therefore,
2g+2f+c = -2 ........ (2)
4g−2f+c = -5 ........(3)
6g+4f+c = -13 ...... (4)
(2) – (3) gives −2g + 4f = 3 ... (5)
(4) – (3) gives 2g + 6f = -8  ....(6)
(5) + (6) gives f = \(-\frac { 1 }{ 2 } \) 
Substituting f = \(-\frac { 1 }{ 2 } \) in (6), g = -\(\frac { 5 }{ 2 } \)
Substituting f = \(-\frac { 1 }{ 2 } \) and g = - \(\frac { 5 }{ 2 } \) in (2), c = 4 .
Therefore the required equation of the circle is
x²+ y²+ 2\(\left( -\frac { 5 }{ 2 } \right) x+2\left( -\frac { 1 }{ 2 } \right) \)y+4 = 0 and x² + y² − 5x − y + 4 = 0.

The discriminant Δ =((4p)² - 4(4)(p+2) = 16(p²-p-2) = 16(p+1)(p-2). So we get
Δ < 0 if -1 < p < 2
Δ = 0 if p = -1 or p = 2
Δ >0 if \(\infty\). Thus the given polynomial has
imaginary roots if -1 < p < 2
equal real roots if p = −1 or p = 2;
distinct real roots if -\(\infty\) < p < -1 or 2 < p < \(\infty\)

x = 1

\(a\left( \frac { 3 }{ 2 } \right) =2\Rightarrow a\frac { 4 }{ 3 } \)
b² = a²(e² - 1)
\(\Rightarrow b^{2}=\frac{16}{9}\left(\frac{9}{4}-1\right)=\frac{16}{9}\left(\frac{9-4}{4}\right)=\frac{\not 16}{9} \times \frac{5}{4}\)
⇒ \({ b }^{ 2 }=\frac { 4\times 5 }{ 9 } =\frac { 20 }{ 9 } \)
∴ Equation of the hyperbola is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
⇒ \(\frac { { x }^{ 2 } }{ \frac { 16 }{ 9 } } -\frac { { y }^{ 2 } }{ \frac { 20 }{ 9 } } =1\)
⇒ \(\frac { { 9x }^{ 2 } }{ 16 } -\frac { 9{ y }^{ 2 } }{ 20 } =1\)

Length of latus rectrum = 8, e = \(\frac35\) and major axis on x-axis 
Given \(\frac { { 2b }^{ 2 } }{ a } =8,e=\frac { 3 }{ 5 } \)
b² = 4a
b² = a²(1- e²)
\(4a={ a }^{ 2 }\left( 1-\frac { 9 }{ 25 } \right) \)
\(4=a\left( \frac { 25-9 }{ 25 } \right) \)
100 = a(16)
a = \(\frac { 100 }{ 16 } =\frac { 25 }{ 4 } \Rightarrow { a }^{ 2 }=\frac { 625 }{ 16 } \)
\({ b }^{ 2 }=4\times \frac { 25 }{ 4 } =25\)
Since major axis is on x-axis, equation of the ellipse is
⇒ \(\frac { { \left( x-0 \right) }^{ 2 } }{ { a }^{ 2 } } +\frac { { \left( y-0 \right) }^{ 2 } }{ { b }^{ 2 } } =1\) 
⇒ \(\frac { { x }^{ 2 } }{ \frac { 625 }{ 16 } } +\frac { { y }^{ 2 } }{ 25 } =1\)
⇒ \(\frac { 16{ x }^{ 2 } }{ 625 } +\frac { { y }^{ 2 } }{ 25 } =1\)

Given \(\frac { dy }{ dx } +y=cosx\)
This is a linear differential equation
Here p = 1, Q = cos x
\(\therefore \int { p\ dx } =\int { dx } =x\)
\(I.F={ e }^{ \int { p\ dx } }={ e }^{ x }\)
The solution is
\({ y }^{ \int { p\ dx } }=\int { Q{ e }^{ \int { p\ dx } }dx+c } \)
\(\Rightarrow { ye }^{ x }=\int { cosx.{ e }^{ x }dx+c } \)
\(\Rightarrow { ye }^{ x }=\frac { { e }^{ x } }{ 2 } \left( cosx+sinx \right) +c\)
\(\Rightarrow y=\frac { 1 }{ 2 } \left( cosx+sinx \right) +{ ce }^{ x }\)
\(\therefore \int { { e }^{ ax }cos\ bx\ dx=\frac { { e }^{ ax } }{ { a }^{ 2 }+{ b }^{ 2 } } \left[ acos\ bx+sin\ ax \right] } \)

Equation of family of parabolas with axis as y axis is given by,
(x-0) = 4a(y-k) .... (1)

Given: Vertex at (0, - 1).
Putting k = -1 in (1), we get
⇒ x² = \(\pm\)4a(y + 1) ....(2)
Differentiating with respect to 'x'  
2x = \(\pm\)4a\(\left( \frac { dy }{ dx } \right) \) ....(3)
⇒ 4a = \(\frac { 2x }{ \frac { dy }{ dx } } \) 
\(\frac{x^2}{2x} = \frac{y+1}{\frac{dt}{dx}}\)
ie) \(x \frac{dy}{dx}-2(y+1) =0\)
This is the required differential equation.

Given the circles centre on r-axis & the circle is passing through the origin.
Let it be (r, 0) & its radius r.

Equation of the circle is
(x - a)² + (y - b)² = r²
(x - r)² + (y - 0)² = r²
⇒ x² - 2xr + r² + y² = r²
⇒ x² - 2xr + y² = 0 ...(1)
defferentiating equation (1) with respect to 'x' we get
⇒ 2x - 2r + 2y \(\frac { dy }{ dx } =0\)
⇒ 2x + 2y \(\frac { dy }{ dx } =2r\)
⇒ x + y \(\frac { dy }{ dx } =r\) ...(2)
Substituting r value in equation (1), we get
x² - 2x \(\left( x+y\frac { dy }{ dx } \right) +{ y }^{ 2 }=0\)
\(\Rightarrow \ { x }^{ 2 }-{ 2x }^{ 2 }-2xy\frac { dy }{ dx } +{ y }^{ 2 }=0\)
\(\Rightarrow \ { -x }^{ 2 }{ -2x }y\left( \frac { dy }{ dx } \right) { +y }^{ 2 }\)
Multiply by '-', we get
\(\Rightarrow \ { x }^{ 2 }{ +2x }y\left( \frac { dy }{ dx } \right) { -y }^{ 2 }\) which is the required differential equation.

Given line is y = x + c
m = 1, c = c
Equation of the hyperbola is 9x² - 16y² = 144
\(\div \)144 we get \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } \) = 1
a² = 16, b² = 9
The condition for the line y = mx + c to be a tangent to the hyperbola is c² = a²m² - b²
∴ c² = 16(1)² - 9 = 16 - 9 = 7
∴ c = ±\(\sqrt7\)

The roots of x³+2x²+3x+4 = 0 are ∝, β, ૪
∴ ∝ + β + ૪ = -co-efficient of x² = -2   .........(1)
∝β + β૪ + ૪∝ = co-effficient of x = 3   ..........(2)
-∝β૪ = +4 ⇒ ∝β૪ = -4  .......(3)
From the cubic equation whose roots are \(\frac { 1 }{ \alpha } ,\frac { 1 }{ \beta } ,\frac { 1 }{ \gamma } \)
\(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } +\frac { 1 }{ \gamma } =\frac { \beta \gamma +\gamma \alpha +\alpha \beta }{ \alpha \beta \gamma } =\frac { 3 }{ -4 } =\frac { -3 }{ 4 } \)
\(\frac { 1 }{ \alpha \beta } +\frac { 1 }{ \beta \gamma } +\frac { 1 }{ \gamma \alpha } =\frac { \gamma +\alpha +\beta }{ \alpha \beta \gamma } =\frac { -2 }{ -4 } =\frac { 1 }{ 2 } \)
\(\left( \frac { 1 }{ \alpha } \right) \left( \frac { 1 }{ \beta } \right) \left( \frac { 1 }{ \gamma } \right) =\frac { 1 }{ \alpha \beta \gamma } =\frac { 1 }{ -4 } =-\frac { 1 }{ 4 } \)
∴ The required cubic equation is 
\({ x }^{ 3 }-\left( \frac { 1 }{ \alpha } +\frac { 1 }{ \beta } +\frac { 1 }{ \gamma } \right) { x }^{ 2 }+\left( \frac { 1 }{ \alpha \beta } +\frac { 1 }{ \beta \gamma } +\frac { 1 }{ \gamma \alpha } \right) x-\left( \frac { 1 }{ \alpha } ,\frac { 1 }{ \beta } ,\frac { 1 }{ \gamma } \right) \)
\(\Rightarrow { x }^{ 3 }+\frac { 3 }{ 4 } { x }^{ 2 }+\frac { 1 }{ 2 } x+\frac { 1 }{ 4 } =0\)
Multiplying by 4 we get, 
4x³ + 3x² + 2x + 1 = 0

2cos²x - 9cosx + 4 = 0 ............ (1)
The left hand side of this equation is not a polynomial in x. But it looks like a polynomial. In fact, we can say that this is a polynomial in cos x. However, we can solve the equation (1) by using our knowledge on polynomial equations. If we replace cos x by y, then we get the polynomial equation 2y²- 9y + 4 = 0 for which 4 and \(\frac{1}{2}\) are solutions.
From this we conclude that x must satisfy cos x = 4 or cos x = \(\frac{1}{2}\).
But cos x = 4 is never possible, if we take cos x = \(\frac{1}{2}\), then we get infinitely many real numbers x satisfying cos x = \(\frac{1}{2}\); in fact, for all n\(\in \)Z, x = 2nπ ±\(\frac { \pi }{ 3 } \) are solutions for the given equation (1).
If we repeat the steps by taking the equation cos²x - 9 cosx + 20 = 0, we observe that this equation has no solution.

Since the circle touch both the axis. Its equation will be
(x + a)² + (y + a)² = a² ...............(1)
It passes through (-4, -2)
∴ (-4 + a)² + (-2 + a)² = a²
\(16+\not a^{2}+8 a+4+a^{2}+4 a=\not a^{2}\)
⇒ a² + 12a + 20 = 0
⇒ (a + 10)(a + 2) = 0
a = -10 or -2

Case (i):
When a = -10, (1) becomes
(x + 10)² + (y + 10)2 = 10²
\(\Rightarrow x^{2}+\not 100+20 x+y^{2}+\not 100+20 y=160\)
⇒  x² + y²+ 20x + 20y + 100 = 0
Case (ii):
When a = -2, (1) becomes
⇒ (x + 2)² + (y + 2)² = 2²
\(x^{2}+4 x+4+y^{2}+4 y+\not 4 = \not 4\)
x² + y²+ 4x + 4y + 4 = 0
Hence, equation of the circles are
x² + y²+ 4x + 4y + 4 = 0
or  x² + y²+ 20x + 20y + 100 = 0