Take the parabolic orbit as open rightward and the vertex at the origin.
Let P be the position of the comet in which FP = 80 million kms.
Draw a perpendicular PQ from P to the axis of the parabola.

Let FQ = x₁
From the triangle FQP,
 \( \sin \frac{\pi}{3} =\frac{P Q}{F P} \)
\(\frac{\sqrt{3}}{2} =\frac{P Q}{80} \)
\(P Q =80 \times \frac{\sqrt{3}}{2}=40 \sqrt{3} \)
\(\cos \frac{\pi}{3} =\frac{F Q}{F P} \)
\(\frac{1}{2} =\frac{x_{1}}{80} \)
x₁ = 40
VQ = a + 40 if VF = a;
P is (VQ, PQ) = (a + 40, 40√3)
Since P lies on the parabola y² = 4ax
(40-√3)² = 4a(a + 40)
⇒ a = -60 or 20
a = -60 is not acceptable
a = 20
The equation of the orbit is 
y² = 4 \(\times\) 20 \(\times\) x
y = 80x
The shortest distance between the Sun and the Comet VF = a = 20 million kms

Let the vertex be at the origin
VF = a = 900
 
The equation of the parabola is y² = 4 \(\times\) 900 \(\times\) x
Let x, be the depth of the mirror at the middle
Since (x₁, 80) lies on the parabola
\(80^{2}=4 \times 900 \times x_{1} \Rightarrow x_{1}=\frac{16}{9}\)
depth of the mirror = \(\frac{16}{9} \mathrm{~cm}\)

Consider the parabolic girder as open downwards
ie., x² = -4ay
it passes through (50, -10)
\( \therefore 50 \times 50 =-4 a(-10) \)
\(\Rightarrow a =\frac{250}{4} \)
\(\therefore x^{2} =-4\left(\frac{250}{4}\right) y \)
\(x^{2} =-250 y \)
Let B(10, y₁) be a point on the parabola
100 = -250y,
\(y_{1}=-\frac{100}{250}=-\frac{2}{5}\)
Let B(10, y₁) be apoint on the parubola
100 = -250y₁
\(y_{1}=-\frac{100}{250}=-\frac{2}{5}\)
Let AB be the height of the bridge at 10 ft. To the right from the midpoint
AC = 10 and BC =
\(A B=10-\frac{2}{5}=9 \frac{3}{5} \mathrm{ft}\)
i.e. The height of the bridge at the required place \(=9 \frac{3}{5} \mathrm{ft}\)

Given foci (± 6, 0)
Length of latus rectum = 10
The transverse axis is along the x axis
Equation of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
foci = (±c, 0) = (± 6,0)
c = 6
Length of Latus rectum = 10
\( \frac{2 b^{2}}{a} =10 \)
\(b^{2} =5 a \)
 \( c^{2}-a^{2} =5 a \)
\(6^{2}-a^{2} =5 a \)
\(36-a^{2} =5 a \)
\(a^{2}+5 a-36 =0 \)
(a + 9)(a - 4) = 0  
a  = -9  a = 4
(not possible)
a = 4 then b² = 5(4) = 20
The required equation of hyperbola
\(\frac{x^{2}}{16}-\frac{y^{2}}{20}=1\)

Given vertices are (9, 2) & (1, 2)
A (9, 2), A' (1, 2)
Centre is midpoint of AA'
\(C(h, k)=\left(\frac{9+1}{2}, \frac{2+2}{2}\right)\)
=(5, 2)
AA' = 2a = 8
a = 4
FF' = 2c = 10
c = 5
\(b^{2}=c^{2}-a^{2}\)
= 25 - 16
b² = 9
The transverse axis is parallel to x axis
Equation of hyperbola \( \frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1 \)
\( \frac{(x-5)^{2}}{16}-\frac{(y-2)^{2}}{5}=1 \)
Length of Latus rectum \( =\frac{2 b^{2}}{a}=\frac{2(9)}{4} \)
\( =\frac{9}{2} \text { units } \)

Given equation of directrix y = \(\pm\)9
foci = (0, \(\pm\)4)
The major axis is along the y axis Equation of the ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\)
foci = c = 4 \(\Rightarrow\) ae = 4 ............. (1)
directrix \(\frac{a}{e}=9\) ......... (2)
\( (1) \times(2) \Rightarrow a \not e \times \frac{a}{\not e} =36 \)
\(a^{2} =36 \)
\(b^{2} =a^{2}-c^{2} \)
\( =36-16 \)
\(b^{2} =20 \)
\(\therefore\) The required Equation of ellipse
\(\frac{x^{2}}{20}+\frac{y^{2}}{36}=1\)
Length of Latus rectum \( =\frac{2 b^{2}}{a} \)
\( =\frac{\not 2(20)}{\not 6_3}=\frac{20}{3} \text { units } \)

The major axis is along the y axis
Equation of ellipse : \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\)
Foci \( =(0, \pm 4) \)
\(c =4 \ \ a e =4 \)
\(a\left(\frac{4}{5}\right) =4 \)
\(a = \not4 \times \frac{5}{\not4} \)
a = 5 
\( b^{2} =a^{2}-c^{2} \)
\(=25-16 b^{2} =9 \)
The required equation of ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{25}=1\)

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 \( =(h, \pm c+k) \)
\(=(1, \pm 2 \sqrt{6}+4) \)
\( =(1,2 \sqrt{6}+4) 8(1,-2 \sqrt{6}+4)\)

\( 4 x^{2}+3 y^{2}+8 x+12 y+4 =0 \)
\(\left(4 x^{2}+8 x\right)+\left(3 y^{2}+12 y\right)+4 =0 \)
\(4\left(x^{2}+2 x\right)+3\left(y^{2}+4 y\right) =-4 \)
\(4\left\{(x+1)^{2}-1\right\}+3\left\{(y+2)^{2}-4\right\} =-4 \)
\(4(x+1)^{2}+3(y+2)^{2} =12 \)
\(\div \text { by } 12 \Rightarrow \quad \frac{(x+1)^{2}}{3}+\frac{(y+2)^{2}}{4} =1 \)
The major axis is parallel to y axis
\( a^{2}=4\ \ b=3 \)
\(a=2 \ \ b=\sqrt{3} \)
\(c^{2}=a^{2}-b^{2} \)
 = 4 - 3 
\(c^{2}=1 \) c = 1
\( a e =1 \Rightarrow 2(e)=1 \)
\(e =\frac{1}{2} \)
Centre (h, k) = (-1 , -2)
Equation of directrix y \( =\pm \frac{a}{e}+k \)
\( =\pm \frac{2}{\frac{1}{2}}-2 \)
\( =\pm 4-2 \)
y = 4 - 2 & y = -4-2
y = 2 & y = -6
Equation of latus rectum
\( \mathrm{y}=\pm \mathrm{ae}+\mathrm{k} \)
\( =\pm(2)\left(\frac{1}{2}\right)-2=\pm 1-2 \)
\( y=1-2 \ \ \quad y=-1-2 \)
\( y=-1 \ \ \quad y=-3 \)
Length of latus rectum \(=\frac{2 b^{2}}{a}=\frac{2(3)}{2}=3\)

\( 6 x^{2}+9 y^{2}+12 x-36 y-12 =0 \)
\(\left(6 x^{2}+12 x\right)+\left(9 y^{2}-36 y\right) =12 \)
\(6\left(x^{2}+2 x\right)+9\left(y^{2}-4 y\right) =12 \)
\(6\left\{(x+1)^{2}-1\right\}+9\left\{(y-2)^{2}-4\right\} =12 \)
\(6(x+1)^{2}+9(y-2)^{2}=12+6 +36 \)
\(6(x+1)^{2}+9(y-2)^{2} =54 \)
\(\frac{(x+1)^{2}}{9}+\frac{(y-2)^{2}}{6} =1 \)
\( X=x+1 ; Y =y-2 \)
The equation becomes \(\frac{X^{2}}{9}+\frac{Y^{2}}{6}=1\)
Clearly the major axis is along X-axis and the minor axis is along Y-axis.
The equation of the major axis Y = 0 and the equation of the minor axis is X = 0.
The equation of the major axis is y - 2 = 0 and of minor axis is x + 1 = 0.
i.e., the equation of the major axis is y - 2 = 0
Here \(a^{2}=9, b^{2}=6 \Rightarrow a=3 b=\sqrt{6}\)
\(\therefore\) The length of major axis (24) - 6
The length of minor axis (25) = 2\(\sqrt 6\)

Let F and F be the fixed points (4,0) and (- 4,0) respectively and P(x, y) be the moving point.
It is given that \(F_{1} P+F_{2} P=10\)
\(\text { i.e., } \sqrt{\left(x_{1}-4\right)^{2}\left(y_{1}-0\right)^{2}}+\sqrt{\left(x_{1}+4\right)^{2}\left(y_{1}-0\right)^{2}}=10\)
Simplifying we get
\( 9 x_{1}^{2}+25 y_{1}^{2}=225 \)
The locus of (x₁, y₁) is \( \frac{x^{2}}{25}+\frac{y^{2}}{9}=1 \)

From the given data since the major axis is parallel to the x axis, the equation is of the form
\(\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1\)
The centre (h, k) is (4, -1)
 \(\frac{(x-4)^{2}}{a^{2}}+\frac{(y+1)^{2}}{b^{2}}=1\)
It passes through (8, 0) \(\therefore \frac{16}{a^{2}}+\frac{1}{b^{2}}=1\) ....... (1)
But CF₁ = ae = 3
\( b^{2} =a^{2}\left(1-e^{2}\right) \)
\( =a^{2}-a^{2} e^{2}=a^{2}-9 \)
\( (1) \Rightarrow \frac{16}{a^{2}}+\frac{1}{a^{2}-9} =1 \)
\(\Rightarrow 16 a^{2}-144+a^{2} =a^{4}-9 a^{2} \)
\(\Rightarrow a^{4}-26 a^{2}+144 =0 \)
\(\Rightarrow a^{2} =18 \text { or } 8\)
Case (1) : \( a^{2}=18 \)
\( b^{2}=a^{2}-9=18-9=9 \)
Case (2) :
\( a^{2}=8 \)
\(b^{2}=8-9=-1 \) which is not possible
\(\therefore a^{2}=18, b^{2}=9\)
Thus the equation is \(\frac{(x-4)^{2}}{18}+\frac{(y+1)^{2}}{9}=1\)

The parabola is open rightwards
4a = 4 \(\Rightarrow\) d = 1
Vertex (h, k) = (-1, 2)
focus = (a + h, 0 + k)
= (1 -1, 0+2) = (0, 2)
Equation of directrix x = -a + h
= -1 -1
= -2
axis y - 2 = 0
y = 2
Length of latus rectum 4a = 4.

Equation of the ellipse is
2x² + 7y² = 14
\(\text { i.e., } \frac{x^{2}}{7}+\frac{y^{2}}{2}=1\)
Here a² = 7, b² = 2
Let the equation of the tangent be
\( y=m x+\sqrt{a^{2} m^{2}+b^{2}} \)
\( \therefore y=m x+\sqrt{7 m^{2}+2} \)
Since this line passes through the point (5, 2)
we get
\( 2 =5 m+\sqrt{7 m^{2}+2} \)
i.e., \( 2-5 m =\sqrt{7 m^{2}+2} \)
(2-5m)² = 7m² + 2
4+25m² - 20m = 7m² +2
18m² - 20m + 2 = 0
9m² -10m + 1 = 0
(9m -1) (m-1) = 0
 \(\therefore m=1 \text { or } m=\frac{1}{9}\)
To find the equations of the tangents, use slope-point from
(i) m = 1,
The equation is
y - y₁ = m(x-x₁).
y - 2 = 1(x - 5)
i.e., x - y - 3 = 0
(ii) \(m=1 / 9\)
The equation is \(y-2=\frac{1}{9}(x-5)\)
i.e., x - 9y + 13 = 0
Thus the equations of the tangents are
x - y - 3 = 0, x - 9y + 13 = 0

Equation of the parabola is y² = 12x
Here 4a = 12 = a = 3
Parametric points at t = \(\left(a t^{2}, 2 a t\right)\)
At t = 1 \(\Rightarrow\) (3(1), 2(3)(1)) = (3, 6)
Equation of tangent
\( y y_{1}=12 \frac{\left(x+x_{1}\right)}{2} \)
At (3, 6)
\( y(6)=\frac{12(x+3)}{2} \text { i.e., } x-y+3=0 \)
Alternative form :
The equation of the tangent at 't' is yt = \(=x+a t^{2}\)
Here 4a = 12 ⇒ a = 3
Also t = 1
The equation of the tangent is y = x + 3
x - y + 3 = 0