\(\frac { { \left( x+3 \right) }^{ 2 } }{ 225 } -\frac { { \left( y-4 \right) }^{ 2 } }{ 64 } =1\)
Given equation is \(\frac { { \left( x+3 \right) }^{ 2 } }{ 225 } -\frac { { \left( y-4 \right) }^{ 2 } }{ 64 } =1\)
This is an equation of the hyperbola
∴ a² = 225, b² = 64
⇒ c² = a² + b² = 225 + 64 = 289
⇒ c = 17
\(e=​​​​\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 64 }{ 225 } } =\sqrt { \frac { 225+64 }{ 225 } } \)
= \(\sqrt { \frac { 289 }{ 225 } } =\frac { 17 }{ 15 } \)
(a) Center is (-3, 4)
⇒ h = -3, k = 4
(b) Foci are (h + c, k), (h - c, k)
⇒ (-3 + 17,4), (-3 -17, 4)
⇒ (14, 4) (-20, 4)
(c) Vertices are (h + a, k) and (h - a, k)
⇒ (-3 + 15,4), (-3 - 15,4)
⇒ (12, 4) (-18, 4)
(d) Equation of directrices are x + 3 = \(\pm \frac { a }{ e } \)
\(\Rightarrow x+3=\pm \frac { 15 }{ \frac { 17 }{ 15 } } \Rightarrow x+3=\pm \frac { -225 }{ 17 } \)
\(\Rightarrow x=\frac { 225 }{ 17 } \) and \(x=\frac { -225 }{ 17 } \)
\(\Rightarrow x=\frac { 225-51 }{ 17 } \) and \(x=\frac { -225-51 }{ 17 } \)
\(\Rightarrow x=\frac { 174 }{ 17 } \) and \(x=\frac { -276 }{ 17 } \)

  \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1\)
Given equation is \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1\)
This is an equation of the hyperbola where the transverse axis is parallel to the y-axis.
∴ a² = 16, b² = 9, c² = a² + b²
⇒ c² = 16 + 9 = 25 ⇒ c = 5
\(e=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1+\frac { 9 }{ 16 } } =\sqrt { \frac { 16+9 }{ 16 } } =\sqrt { \frac { 25 }{ 16 } } =\frac { 5 }{ 4 } \)
(a) Center is (0, 0)
⇒ h = 0, k = 0

(b) Vertices are (h, k + a), (h, k-a)
⇒ (0, 0 + 4), (0, 0 - 4) ⇒ (0, 4) (0,-4)
(c) Foci are (h, k + c), (h, k- c)
⇒ (0, 0 + 5), (0,  0 - 5) ⇒ (0, 5) (0, -5)
(d) Equations of Directrices are y = \(x=\pm \frac { a }{ e } \)
\(\Rightarrow y=\pm \frac { 4 }{ \frac { 5 }{ 4 } } \Rightarrow y=\pm \frac { 16 }{ 5 } \)

\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 144 } =1\)

This is an equation of the hyperbola.
∴ a² = 25 and b² = 144
⇒ c² =a² + b² =25 + 144 =169 ⇒ c = 13
\(e=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1+\frac { 144 }{ 25 } } =\sqrt { \frac { 169 }{ 25 } } =\frac { 13 }{ 5 } \)
(a) Center is (0, 0) ⇒ h = 0, k = 0
(b) Foci are (h + c, k), (h - c, k)
⇒ (0 + 13,0), (0 - 13,0)
⇒ (13, 0), (-13, 0)
(c) Vertices are (h + a, k) and (h - a, k)
⇒ (0 + 5, 0), (0 - 5, 0) ⇒ (5, 0), (-5, 0)
(d) Equations of Directrices are x = \(x=\pm \frac { a }{ e } \)
\(\Rightarrow x=\frac { 5 }{ \frac { 13 }{ 5 } } \Rightarrow x=\pm \frac { 25 }{ 13 } \)

Since the truck’s width is 3m, to determine the clearance, we must find the height of the archway 1.5 m from the centre. If this height is 2.7 m or less the truck will not clear the archway.

From the diagram a = 6 and b = 3 yielding the equation of ellipse as \(\frac { { x }^{ 2 } }{ { 6 }^{ 2 } } +\frac { { y }^{ 2 } }{ { 3 }^{ 2 } } =1\)
The edge of the 3m wide truck corresponds to x = 1.5 m. We will find the height of the
archway 1.5 m from the centre by substituting x =1.5 and solving for y 
\(\frac { { \left( \frac { 3 }{ 2 } \right) }^{ 2 } }{ 36 } \frac { { y }^{ 2 } }{ 9 } =1\)
 \({ y }^{ 2 }=9\left( 1-\frac { 9 }{ 144 } \right) \)
\(\frac { 9\left( 135 \right) }{ 144 } =\frac { 135 }{ 16 } \)
\(y=\frac { \sqrt { 135 } }{ 4 } \)
= 2.90
Thus the height of arch way 1.5m from the centre is approximately 2.90m. Since the truck’s height is 2.7 m, the truck will clear the archway.

The parametric equation of tangent at 't₁' to the parabola y² = 4ax is yt₁ = x + at₁²  ...(1)
Also, the parametric equation of tangent at 't₂' to the parabola y = 4ax is yt₂ = x+ at₂² ...(2)
(1) ➝ yt₁ = x + at₁²
(2) ➝ yt₂ = x + at₂²
(1) - (2) y(t₁ -t₂) = a(t₁² - t₂²)

⇒ y = a(t₁ + t₂)
Substitutingy = a(t₁ + t₂) in (1) we get,
a(t₁ + t₂)t₁ = x + at₁²

⇒ x = at₁t₂
Hence, the point of intersection of two lengths is
[at₁t₂, a(t₁ + t₂)]

\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\)
This is an equation of the ellipse.
a² = 2⁵ and b² = 9 and c² = a² - b²
⇒ c² = 25 - 9 = 16 ⇒ c = 4
(a) Center is (0, 0) ⇒ h = 0, k = 0
(b) foci are (h - c, k), (h + c, k)
⇒ (0 - 4, 0), (0 + 4, 0)
⇒ (-4, 0) and (4, 0)
(c) Vertices are (h - a, k) and (h + a, k)
⇒ (0 - 5, 0) and (0 + 5, 0)
⇒ (-5, 0) and (5, 0)
(d) Directrices are x = \(\pm \frac { a }{ e } \)
⇒ x = \(\pm \frac { 5 }{ e } \)
\(e=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 9 }{ 25 } } =\sqrt { \frac { 16 }{ 25 } } =\frac { 4 }{ 5 } \)
∴ Directrice are x = \(\pm \frac { 5 }{ \frac { 4 }{ 5 } } \Rightarrow x=\pm \frac { 25 }{ 4 } \)

Let the vertex be (0, 0) .
The equation of the parabola is
y² = 4ax
(1) Since the diameter is 40cm and the depth is 30 cm , the point (30, 20) lies on the parabola.
20² = 4a × 30
4a = \(\frac { 400 }{ 30 } \) = \(\frac { 40 }{ 3 } \)
Equation is y² = \(\frac { 40 }{ 3 } \)x.
(2) The bulb is at focus (0, a).
Hence the bulb is at a distance of \(\frac { 10 }{ 3 } \)cm from the vertex.

Equation of the parabola is y = \(\frac { 1 }{ 32 } \)x²
That is x² = 32y ; the vertex is (0, 0)
= 4 (8)y
\(\Rightarrow a=8\)
So the heating tube needs to be placed at focus (0, a) 
Hence the heating tube needs to be placed 8 units above the vertex of the parabola.

Let P(x, y) be any point on the hyperbola
\(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)

Then by definition, SP = ePM and S'P = ePM'
SP = ePM ⇒ SP = e(NK)
= e(CN - CK)
= \(e\left( x-\frac { a }{ e } \right) \)  = ex - a
= and S'P = ePM' ⇒ S'P = e(NK')
⇒ e(CN + CK') = \(e\left( x+\frac { a }{ e } \right) \) = ex + a
∴ S'P -SP = (ex + a) - (ex - a)
ex + a - ex + a = 2a (constant)
= length of the transverse axis.

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.