$9x^2+5y^2=45$

Given ellipse may be written as, $\displaystyle \frac{x^2}{32}+\frac{y^2}{18}=1$
$\Rightarrow a^2 = 32, b^2 = 18$
Hence required normal at $(4,3)$ is given by,
$\displaystyle y-3=\frac{3\times 32}{4\times 18}(x-4)\Rightarrow y-3=\frac{4}{3}(x-4)\Rightarrow 4x-3y=7$

Ellipse:$\cfrac { { x }^{ 2 } }{ a^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b^{ 2 } } } =$

If $y=mx+c$ is normal to ellipse
${ c }^{ 2 }={ m }^{ 2 }\cfrac { { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 }{ m }^{ 2 } } $
${ x }^{ 2 }+3{ y }^{ 2 }=3$
$\cfrac { { x }^{ 2 } }{ 3 } +\cfrac { { y }^{ 2 } }{ 12 } =1$
${ a }^{ 2 }=3,b=1$
$y+3x=c$
$m=-3$
${ c }^{ 2 }={ (-3) }^{ 2 }\cfrac { { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 }{ m }^{ 2 } } =9\times \cfrac { { (3-1) }^{ 2 } }{ 3+9 } $
$=\cfrac { 9\times 4 }{ 12 } =3$

The equation of the normal to the given ellipse at the point $P(a  \cos  \theta,  b   \sin  \theta)$ is $ax  \sec  \theta - by  \cos ec \theta = a^2 - b^2$.
$\Rightarrow    \displaystyle y = \left ( \frac{a}{b} \tan  \theta \right)  x - \frac{(a^2 - b^2)}{b} \sin \theta$      (i)
Let      $\displaystyle \frac{a}{b} \tan  \theta = m$, so that
$\displaystyle \sin \theta = \frac{bm}{\sqrt{a^2 + b^2 m^2}}$
Hence, the equation of the normal Equation (i) becomes
$ y = mx - \displaystyle \frac{(a^2 - b^2)m}{\sqrt{a^2 + b^2 m^2}}$
$\therefore    m  \in  R,$ as $m  = \dfrac{a}{b} tan  \theta  \in  R.$

Let $a>b$, then one of latus rectum of the ellipse is $(ae, \cfrac{b^2}{a})$

Let $a>b$, then one of latus rectum of the ellipse is $(ae, \cfrac{b^2}{a})$
Thus equation of normal at this point is given by,
$\cfrac{a^2x}{ae}-\cfrac{b^2y}{b^2/a}=a^2e^2$
Given it passes through one of minor axis ,which is $(0,-b)$
$\Rightarrow \cfrac{a^2(0)}{ae}+\cfrac{b^2(-b)}{b^2/a}=a^2e^2$
$\Rightarrow e^2=\cfrac{b}{a}$
Now using $e^2=1-\cfrac{b^2}{a^2}$
we get,  $e^4+e^2-1=0$

Given ellipse may be written as, $\displaystyle \cfrac{x^{2}}{16} + \cfrac{y^{2}}{4} = 1$
$\Rightarrow a^2=16, b^2=4, \therefore e=\sqrt{1-\dfrac14}=\dfrac{\sqrt{3}}2$
Then latus rectum of the ellipse in the first quadrant is $(ae, \cfrac{b^2}{a})\equiv (2\sqrt{3},1)$

Equation of tangent to ${ y }^{ 2 }=4x$ at $\left( { t }^{ 2 },2t \right) $ is $x=ty-{ t }^{ 2 }$   ....(1)

Given ellipse is, $\displaystyle \cfrac{x^{2}}{a^2} + \cfrac{y^{2}}{b^2} = 1$
Thus general equation of normal to ellipse with slope $m$ is given by,
$y = mx-\cfrac{(a^2-b^2)m}{\sqrt{a^2+b^2m^2}}$
Given external point through which this line is passing is $P(0,c)$
$\Rightarrow c = -\cfrac{(a^2-b^2)m}{\sqrt{a^2+b^2m^2}}$
$\Rightarrow c^2=\cfrac{(a^2-b^2)^2m^2}{a^2+b^2m^2}$
$\Rightarrow m^2=\cfrac{c^2a^2}{(a^2-b^2)^2-c^2b^2}$

Given ellipse may be written as, $\displaystyle \cfrac{x^{2}}{9} + \cfrac{y^{2}}{4} = 1$
$\Rightarrow a^2=9, b^2=4$
Thus general equation of normal to ellipse with slope $m$ is given by,
$y = mx-\cfrac{(a^2-b^2)m}{\sqrt{a^2+b^2m^2}}=mx-\cfrac{5m}{\sqrt{9+4m^2}}$
Let any external point through wich this line is passing is $P(x _1,y _1)$
$\Rightarrow (y _1-mx _1)^2=\cfrac{25m^2}{9+4m^2}$
$\Rightarrow (y _1-mx _1)^2(9+4m^2)=25m^2$
Clearly this is a polynomial of degree four so maximum number of normal that can be drawn from point $P$ (any external point) to the ellipse is $4$ corresponding to four roots of $m.$

Equation of normal is $\dfrac{a^2x}{x _1}-\dfrac{b^2y}{y _1}=a^2-b^2$

The equation of the normal at $\left( a\cos { \phi ,b\sin { \phi  }  } 

\right) $ to the ellipse $\cfrac { { x }^{ 2 } }{ { a }^{ 2 } }

+\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ is
$ax\sec { \phi +by cosec\phi=\left( { a }^{ 2 }-{ b }^{ 2 } \right)  }  \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$
and the equation of the line is
$lx+my=n\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$
Now $(i)$ and $(ii)$ represent the same line
$\therefore

\quad \cfrac { a\sec { \phi  }  }{ l } =\cfrac { b cosec\phi }{ m } =\cfrac {

\left( { a }^{ 2 }-{ b }^{ 2 } \right) }{ n } $
$\Rightarrow

\sin { \phi  } =\cfrac { bn }{ m(a^2-b^2) } \quad \cos { \phi  } =\cfrac { {

\left( { an } \right)  } }{ l(a^2-b^2) } $
Squaring and adding we get
$ \cfrac { { a }^{ 2 } }{ { l }^{ 2 } } +\cfrac { { b }^{ 2 } }{ { m }^{ 2

} } =\cfrac { { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 } }{ { n

}^{ 2 } } $
Hence. option 'C' is correct.

$\displaystyle P\left ( \frac{1}{2}+, 2 \right ), Q(1, 1)$ 
Tangents at P and Q are $\displaystyle 4x\cdot \frac{1}{2}+y\cdot 2=5$ and $\displaystyle 4x\cdot 1+y\cdot 1=5$ or $\displaystyle 2x+2y=5$ and $\displaystyle 4x+y=5$ 
Hence normals are $\displaystyle 2x-2y+3=0,$ $\displaystyle x-4y+3=0$
 They intersect at $\displaystyle \left ( -1, \frac{1}{2} \right )$

$7\sqrt{3}=\dfrac{42m}{\sqrt{24-18m^2}}\Rightarrow \sqrt{3}=\dfrac{\sqrt{6}m}{\sqrt{4-3m^2}}\Rightarrow 4-3m^2=2m^2$
$m=\dfrac{2}{\sqrt{5}}$.

Given Ellipse $\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 4 } =1$

$e=\sqrt{1-\dfrac{b^2}{a^2}} =\dfrac {\sqrt{3}}{2}$

$\because P$ is a point on the ellipse 

So, $P=\left( 4\cos { \theta  } ,2\sin { \theta  }  \right) $

Equation of normal to the ellipse $\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 4 } =1$ at point $(x _{1},y _{1})=(4\cos \theta, 2\sin \theta)$ is given by

$a^2 y _1(x-x _1)=b^2 x _1(y-y _1)$ 

$\implies 16\times 2\sin \theta(x-4\cos \theta)=4\times 4\cos \theta(y-2\sin \theta)$

$\implies 2x\sin \theta-8\sin \theta \cos \theta=y\cos \theta-2\sin \theta \cos \theta$

$\implies 2x\sin \theta=y\cos \theta + 6\sin \theta \cos \theta$

$\implies \dfrac{2x}{\cos \theta}=\dfrac{y}{\sin \theta}+6$

$\implies 2x\sec \theta -y\text{cosec} \theta = 6$

It meet the x-axis at $Q(3\cos \theta, 0)$

$\therefore M=\left( \dfrac { 7 }{ 2 } \cos { \theta  } ,\sin { \theta  }  \right) =\left( x,y \right) $

Locus of $M$ is

$\dfrac { { x }^{ 2 } }{ { \left( \dfrac { 7 }{ 2 }  \right)  }^{ 2 } } +\dfrac { { y }^{ 2 } }{ 1 } =1$

The equation of the normal to the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ at the point $P(a \cos \theta, b \sin \theta)$ is $ax\sec\theta - bycosec\theta= a^2 - b^2$

Given equation of ellipse is $\dfrac {x^2}{4}+\dfrac {y^2}{9}=1$

Let the feet of normal be $P(x,y)$

Normal is parallel to $2x-y=1$

Therefore, slope of normal at $P$ $=2$

and slope of slope of tangent at $P$ $=-\dfrac { 1 }{ 2 } $

Point of contact of tangent in slope from is $\left( \dfrac { \pm { a }^{ 2 }m }{ \sqrt { { a }^{ 2 }{ m }^{ 2 }+{ b }^{ 2 } }  } ,\dfrac { { \mp b }^{ 2 } }{ \sqrt { { a }^{ 2 }{ m }^{ 2 }+{ b }^{ 2 } }  }  \right) $

Here $a=2,b=3$ and $m=-\dfrac{1}{2}$

Therefore, the point of contact are:

$\left( \dfrac { \pm \left( 4\times \dfrac { -1 }{ 2 }  \right)  }{ \sqrt { 4\times \dfrac { 1 }{ 4 } +9 }  } ,\dfrac { \mp 9 }{ \sqrt { 4\times \dfrac { 1 }{ 4 } +9 }  }  \right) \\ \left( \dfrac { \mp 2 }{ \sqrt { 10 }  } ,\dfrac { \mp 9 }{ \sqrt { 10 }  }  \right) $

 So, option C is correct.

$Equation\quad of\quad ellipse:\quad \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\ Co-ordinates\quad of\quad positive\quad latus\quad rectum\quad is:\quad (ae,\frac { { b }^{ 2 } }{ a } )\ Equation\quad of\quad normal\quad at\quad point(x1,y1)\quad is:\ \frac { { a }^{ 2 }x }{ x1 } -\frac { { b }^{ 2 }y }{ y1 } ={ (ae) }^{ 2 }\ \therefore \quad Equation\quad is:\quad \frac { { a }^{ 2 }x }{ ae } -\frac { { b }^{ 2 }y }{ \frac { { b }^{ 2 } }{ a }  } ={ (ae) }^{ 2 }\ Or,\quad \frac { x }{ e } -\frac { y }{ 1 } ={ ae }^{ 2 }\ Or,\quad x-ey-{ ae }^{ 3 }=0$

Given equation of ellipse $\displaystyle \frac{x^{2}}{18}+\frac{y^{2}}{8}=1$
$\displaystyle \frac {dy}{dx}=\displaystyle \frac {-4x}{9y}$
Slope of tangent to ellipse at $(3,2)$ is 
$m=\displaystyle \frac{-2}{3}$

Equation of tangent to ellipse is 
$y-2=-\displaystyle \frac{2}{3}(x-3)$
$\Rightarrow 2x+3y=12$
Since , the tangent intersect x-axis i.e. $y=0$
$\Rightarrow x=6$
So, tangent intersects x-axis at $(6,0)$

Equation of normal to ellipse is 
$y-2=\displaystyle \frac{3}{2}(x-3)$
$\Rightarrow 3x-2y=5$
Since , the tangent intersect x-axis i.e. $y=0$
$\Rightarrow x=\displaystyle \frac{5}{3}$
So, normal intersects x-axis at $\left(\displaystyle \frac{5}{3} ,0\right)$

So, area of triangle $=\displaystyle \frac { 1 }{ 2 } \begin{vmatrix} 3 & 2 & 1 \ 6 & 0 & 1 \ \frac { 5 }{ 3 }  & 0 & 1 \end{vmatrix}$
$=\displaystyle \frac{13}{3}$ sq.units

Reason is correct.
Equation of normal in parametric form: $\dfrac{ax}{\cos \theta}-\dfrac{by}{\sin \theta}=a^2-b^2$
$\Rightarrow 5\sqrt 2 x-3\sqrt 2 y=25-9=16$
Therefore, assertion is correct but reason is not the correct explanation 

The equation of any normal to $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ is 
       $ax\sec { \phi  } -by\csc { \phi  } ={ a }^{ 2 }-{ b }^{ 2 }$              ......(i)
The straight line $x\cos { \alpha  } +y\sin { \alpha  } =p$ will be a normal to the ellipse $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$, if equation (i) and $x\cos { \alpha  } +y\sin { \alpha  } =p$ represent the same line.
$\therefore \dfrac { a\sec { \phi  }  }{ \cos { \alpha  }  } =\dfrac { -b\csc { \phi  }  }{ \sin { \alpha  }  } =\dfrac { { a }^{ 2 }-{ b }^{ 2 } }{ p } $
$\Rightarrow \cos { \phi  } =\dfrac { ap }{ \left( { a }^{ 2 }-{ b }^{ 2 } \right) \cos { \alpha  }  } $
$\sin { \phi  } =\dfrac { -bp }{ \left( { a }^{ 2 }-{ b }^{ 2 } \right) \sin { \alpha  }  } $
$\because \sin ^{ 2 }{ \phi  } +\cos ^{ 2 }{ \phi  } =1$
$\Rightarrow \dfrac { { b }^{ 2 }{ p }^{ 2 } }{ { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 }\sin ^{ 2 }{ \alpha  }  } +\dfrac { { a }^{ 2 }{ p }^{ 2 } }{ { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 }\cos ^{ 2 }{ \alpha  }  } =1$
$\Rightarrow { p }^{ 2 }\left( { b }^{ 2 }\csc ^{ 2 }{ \alpha  } +{ a }^{ 2 }\sec ^{ 2 }{ \alpha  }  \right) ={ \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 }$

Let $P\equiv (a\cos\theta, b\sin\theta)$
Thus equation of normal to the given ellipse at 'P' is given by,
$ax\sec\theta-by cosec\theta=a^2-b^2$
$\therefore G \equiv ((a-\cfrac{b^2}{a})\cos\theta,0), g\equiv (0,(b-\cfrac{a^2}{b})\sin\theta)$
Thus $PG = \sqrt{\cfrac{b^4}{a^2}\cos^2\theta+b^2\sin^2\theta}=\cfrac{b}{a}\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}$
and $Pg = \sqrt{a^2\cos^2\theta+\cfrac{a^4}{b^2}\sin^2\theta}=\cfrac{a}{b}\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}$
$\therefore PG:Pg = \cfrac{b^2}{a^2} $

Let $P\left( { x } _{ 1 },{ y } _{ 1 } \right) $ be the point of intersection of the
Line $lx+my+n=0$ and the ellipse $\cfrac { { x }^{ 2 }

}{ {a }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$
Then the equation of tangent at $P$ is
$\cfrac

{ { xx } _{ 1 } }{ { a }^{ 2 } } +\cfrac { { yy } _{ 1 } }{ { b }^{ 2 } }

=1\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$
Since $\left( { x } _{ 1 },{ y } _{ 1 } \right) $  is the point of intersection of the line
$lx+my+n=0\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$
Clearly $(i)$ and $(ii)$ represent the same line. Therefore,
$\therefore \quad \cfrac { { x } _{ 1 } }{ { a }^{ 2 }l } =\cfrac { { y } _{ 1 } }{ { b }^{ 2 }m } =\cfrac { 1 }{ -n } $
${ x } _{ 1 }=\cfrac { { -a }^{ 2 }l }{ n } ,\quad { y } _{ 1 }=\cfrac { -{ b }^{ 2 }m }{ n } $
Therefore, the point of intersection of given line and  the given ellipse is
$\left(- \cfrac { { a }^{ 2 }l }{ n } ,\cfrac { { b }^{ 2 }m }{ n }  \right) \quad $
Hence, option 'B' is correct.

Solution:

The equation of tangent to ellipse is $bx cos\theta+ay sin\theta=ab$