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
⇒ \(\frac { { \left( x-0 \right) }^{ 2 } }{ { a }^{ 2 } } +\frac { { \left( y-0 \right) }^{ 2 } }{ { b }^{ 2 } } =1\) 
⇒ \(\frac { { x }^{ 2 } }{ \frac { 625 }{ 16 } } +\frac { { y }^{ 2 } }{ 25 } =1\)
⇒ \(\frac { 16{ x }^{ 2 } }{ 625 } +\frac { { y }^{ 2 } }{ 25 } =1\)

Foci (0, ±4) and end points of major axis
are (0, ±5)
Since foci are (0, ±be) ⇒ be = 4
End points of major axis are (0, ±5)
⇒ b = 5
∴ 5(e) = 4
⇒ e = \(\frac { 4 }{ 5 } \)
Also, a² = b²(1 - e²)
a² = 25\(\left( 1-\frac { 16 }{ 25 } \right) =25\left( \frac { 25-16 }{ 25 } \right) \)
⇒ a² = 9
Equation of the ellipse is \(\frac { { x }^{ 2 } }{ a ^2 }+\frac { { y }^{ 2 } }{ b^2 } =1\)
⇒ \(\frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 25 } =1\)

foci (±3, 0), e = \(\frac { 1 }{ 2 } \)
Since foci are (±ae, 0)
⇒ ae = 3
⇒ a \(\frac { 1 }{ 2 } \) = 3
⇒ a = 6
Centre is (0, 0)
and b² = a² (1 - e²)
⇒  b² = 36\(\left( 1-\frac { 1 }{ 4 } \right) \)
⇒  \({ b }^{ 2 }=36\left( \frac { 3 }{ 4 } \right) \)
⇒ b² = 27
∴ Equation of the ellipse is \(\frac { { x }^{ 2 } }{ 36 } +\frac { { y }^{ 2 } }{ 27 } =1\) 
\(\frac { { x }^{ 2 } }{ 36 } +\frac { { y }^{ 2 } }{ 27 } =1\)

y² = 4ax is the given parabola. Let (\(\alpha\), \(\beta\)) be the given point.
Equation of the normal in parametric form is
y = –tx + 2at + at³        ...(1)
If m is the slope of the normal then m = −t 
Therefore the equation (1) becomes y = mx - 2am- am³ 
Let it passes through (\(\alpha\), \(\beta\)), then
b = ma - 2am- am³
am³ + (2a −\(\alpha\))m+ \(\beta\) = 0
which being a cubic equation in m, has three values of m. Consequently three normals, in general, can be drawn from a point to the parabola, since complex roots of real equation, always occur in conjugate pairs and (1) being an odd degree equation, it has atleast one real root. Hence atleast one normal to the parabola is real.

Let P(x, y) be a point on the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Draw MM′ through P, perpendicular to directrices l and l′
Draw PN ⊥ to x -axis
By definition SP = ePM
= eNZ
= e[CZ -CN]
\(=e\left[\frac{a}{2}-x\right]=a-e x\)      ....(1)
and SP′ = ePM′
= e[CN +CZ']
\(=e\left[x+\frac{a}{e}\right]=e x+a\)    ....(2)
Hence, SP + S'P = a − ex + a + ex = 2a

Let P(x₁, y₁) be a point outside the circle. The equation of the tangent is
\(y=m x \pm a \sqrt{1+m^{2}}\) It passes through (x₁, y₁ ). Therefore
\( y_{1} =m x_{1} \pm a \sqrt{1+m^{2}} \)
\(y_{1}-m x_{1} =a \sqrt{1+m^{2}} . \) Squaring both sides, we get
\( \left(y_{1}-m x_{1}\right)^{2} =a^{2}\left(1+m^{2}\right) \)
\(y_{1}^{2}+m^{2} x_{1}^{2}-2 m x_{1} y_{1}-a^{2}-a^{2} m^{2} =0 \)
\(m^{2}\left(x_{1}^{2}-a^{2}\right)-2 m x_{1} y_{1}+\left(y_{1}^{2}-a^{2}\right) =0\) 
This quadratic equation in m gives two values for m 
These values give two tangents to the circle x² + y² = a² .

Equation of the circle is x² + y² + 2gx + 2 fy + c = 0 with centre C (-g, - f ) and radius \(r=\sqrt{g^{2}+f^{2}-c}\) 
Let P(x₁, y₁) be a point in the plane. Join CP and let it meet the circle at Q. Then the point P is outside, on or within the circle according as
\(|C P| \text { is } \begin{cases}>|C Q| & \text { or, } \\ =|C Q| & \text { or, } \\ <|C Q| . & \end{cases}\)
\(\Rightarrow \quad C P^{2} \text { is } \begin{cases}>r^{2} & \text { or, } \\ =r^{2} & \text { or } \quad\{C Q=r\}, \\ <r^{2} .\end{cases}\)
\(\Rightarrow\left(x_{1}+g\right)^{2}+\left(y_{1}+f\right)^{2} \text { is } \begin{cases}>g^{2}+f^{2}-c & \text { or, } \\ =g^{2}+f^{2}-c & \text { or, } \\ <g^{2}+f^{2}-c\end{cases}\)
\(\Rightarrow \quad x_{1}^{2}+y_{1}^{2}+2 g x_{1}+2 f y_{1}+c \text { is } \begin{cases}>0 & \text { or, } \\ =0 & \text { or, } \\ <0 & \end{cases}\)

Let A (x₁, y₁) and B (x₂, y₂) be the two extremities of the diameter AB and P(x, y) be any point on the circle. Then \(\angle A P B=\frac{\pi}{2}\) (angle in a semi-circle).
Therefore, the product of slopes of AP and PB is equal to -1.
That is, \(\left(\frac{\left(y-y_{1}\right)}{\left(x-x_{1}\right)}\right)\left(\frac{\left(y-y_{2}\right)}{\left(x-x_{2}\right)}\right)=-1\) yielding the equation of the required circle as (x − x ₁)(x − x₂ ) + ( y − y₁ )( y − y2 ) = 0

Let the circle be S : x² + y² +2gx+2 fy+c = 0 , … (1)
and the line be L : lx+my+n = 0 . … (2)
Consider S +\(\lambda\) L = 0 . That is x² + y² + 2gx + 2 fy + c +\(\lambda\) (lx + my + n) = 0    ....(3)
Grouping the terms of x, y and constants, we get \(x^{2}+y^{2}+x(2 g+\lambda l)+y(2 f+\lambda m)+c+\lambda n=0\) which is a second degree equation in x and y with coefficients of x² and y² equal and there is no xy term.
If (\(\alpha\), \(\beta\) ) is a point of intersection of S and L satisfying equation (1) and (2), then it satisfies equation (3).
Hence S + \(\lambda\)L = 0 represents the required circle.

The equation of the ellipse is \(\frac { { \left( x-11 \right) }^{ 2 } }{ 484 } +\frac { { y }^{ 2 } }{ 64 } =1\). The origin of the sound wave and the kidney stone of patient should be at the foci in order to crush the stones.
a² = 484 and b² = 64
c² = a² -b²
= 484-64
= 420
c \(\simeq \) 20.5
Therefore the patient’s kidney stone should be placed 20.5 cm from the centre of the ellipse.

In the given ellipse a² = 16, b² = 9
then c² = a² - b²
c² = 16 - 9
= 7
c = ±\(\sqrt { 7 } \)
Therefore the foci are F(\(\sqrt { 7 } \), 0) F¹(-\(\sqrt { 7 } \), 0). The focus of the parabola is (\(\sqrt { 7 } \), 0) \(\Rightarrow \) a = \(\sqrt { 7 } \)
Equation of the parabola is y² = 4\(\sqrt { 7 } \) x.

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 the parabola be y² = 4ax 
Since focus is 2m from the vertex a = 2
Equation of the parabola is y² = 8x
Let P be a point on the parabola whose x -coordinate is 3m from the
vertex P (3, y) 
y² = 8 × 3
y =\(\sqrt { 8\times 3 } \)
= \(2\sqrt { 6 } \)
The width of the antenna 3m from the vertex is 4\(\sqrt { 6 } \) m.

From the graph the vertex is at (0, 0) and the parabola is open down
Equation of the parabola is x² = -4ay
(-20, -15) and (20, -15) lie on the parabola 
20² = -4a(-15)
\(4a=\frac { 400 }{ 15 } \)
x² =\(\frac { -80 }{ 3 } \) x y
Therefore equation is 3x² = -80y

Length of latus rectum = 4,
distance between foci = 4\(\sqrt { 2 } \) major axis is y-axis
Given \(\frac { { 2b }^{ 2 } }{ a } =4\) and distance between foci = 2ae = 4\(\sqrt { 2 } \)
⇒ ae = \(2\sqrt { 2 } \)
⇒a²e² = 8 ....(1)
\(\frac { { 2b }^{ 2 } }{ a } =4\Rightarrow { b }^{ 2 }=2a\) ...(2)
We know b² = a²(1 - e²)
b² = a² - a²e²
2a = a2 - 8
[using (1) and (2)]
a² - 2a - 8 = 0
On factorising we get
(a - 4)(a + 2) = 0
a = 4 or-2
a = 4
⇒ [∴ a = -2 is not possible]
⇒ ae = 16
∴ From (2), b² = 2(4) = 8
Hence, the equation of the ellipse is
\(\frac { { x }^{ 2 } }{ 8 } +\frac { { y }^{ 2 } }{ 16 } =1\) [∵ Major axis is y -axis]

From figure the midpoint of line joining foci is the centre C(0, 0).
Transverse axis is y-axis 
AA′ = 2a \(\Rightarrow \) 2a = 8,
SS′ = 2c = 12, c = 6
a = 4
b² = c²−a² = 36−16 = 20
Hence the equation of the required hyperbola is \(\frac { { y }^{ 2 } }{ 16 }- \frac { { x }^{ 2 } }{ 20 } =1\)

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 

C(2, 1) is the centre and 3x + 4y + 10 = 0 cuts a chord AB on the circle. Let M be the midpoint of AB,
then AM = BM = 3. Now BMC is a right triangle.
So, we have CM = \(\frac { \left| 3\left( 2 \right) +4\left( 1 \right) +10 \right| }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } } =4\)
By Pythagoras theorem BC² = BM² + MC² = 3² + 4² = 25
BC = 5 = radius
Equation of the required circle is
(x−2)²+(y−1) = 5²
x²+y²−4x−2y−20 = 0 .

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.

The Latus rectum LL′ of an ellipse \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\) passes through S(ae, 0)  
Hence L is (ae, y₁ )
Therefore, \(\frac { { a }^{ 2 }{ e }^{ 2 } }{ { a }^{ 2 } } +\frac { { y_1 }^{ 2 } }{ { b }^{ 2 } } \) = 1
\(\frac { { y _1}^{ 2 } }{ { b }^{ 2 } } \) = 1-e²
y₁²  = b²(1-e²)
= b²\(\left( \frac { { b }^{ 2 } }{ { a }^{ 2 } } \right) \)\(\left( since,{ e }^{ 2 }=1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } \right) \ \)
y₁ = \(\pm\frac { { b }^{ 2 } }{ { a } } \)
That is the end points of Latus rectum L and L′ are \(\left( ae,\frac { { b }^{ 2 } }{ { a } } \right) and\ \left( ae-\frac { { b }^{ 2 } }{ { a } } \right) \)
Hence the length of latus rectum LL' = \(\frac { { b }^{ 2 } }{ { a } } \)