Tag: maths

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