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Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The number of normals to the ellipse $\dfrac { { x }^{ 2 } }{ 25 } +\dfrac { { y }^{ 2 } }{ 16 } =1$ which are tangents to the circle ${ x }^{ 2 }+{ y }^{ 2 }=9$ is

  1. 1

  2. 2

  3. 3

  4. 0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A normal to an ellipse is tangent to a circle if the distance from the center to the normal equals the radius. For the given ellipse and circle, there is only one such normal.

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The equation of the normal to the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ at the end of latus rectum in quadrant $1^{st}$ and $4^{th}$ is

  1. $x-ey-ae^3=0$
  2. $x+ey-ae^3=0$
  3. $y-ex-be^3=0$
  4. $y+ex-be^3=0$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

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

ends of L.R in 1st and 4th quadrant is $L(ae,\cfrac { b^{ 2 } }{ a } )$ and $L'(ae,\cfrac { -b^{ 2 } }{ a } )$
equation of normal at $L$ and $L'$
at $L$, $\Rightarrow \cfrac { a^{ 2 }x }{ ae } -\cfrac { b^{ 2 }y }{ \cfrac { b^{ 2 } }{ a }  } =a^{ 2 }-b^{ 2 }$
$ \Rightarrow ax-aey=e(a^{ 2 }-b^{ 2 })$
$ \Rightarrow x-ey=\cfrac { a^{ 2 } }{ a } (e^{ 2 })(e)$
$ \Rightarrow x-ey-ae^{ 3 }=0---(i)$
at $L'$ $\Rightarrow \cfrac { a^{ 2 }x }{ ae } -\cfrac { b^{ 2 }y }{ -\cfrac { b^{ 2 } }{ a }  } =a^{ 2 }e^{ 2 }$
$\Rightarrow x+ey-ae^{ 3 }=0---(ii)$

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

If line $y+3x=c$ is normal of the ellipse ${ x }^{ 2 }+3{ y }^{ 2 }=3$ then equation of normal is-

  1. $y-3x\pm \sqrt { 3 } =0$
  2. $y+3x\pm \sqrt { 3 } =0$
  3. $y+3x\pm 3 =0$
  4. $y+3x\pm 1 =0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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 } $
$=\cfrac { 9\times 4 }{ 12 } =3$

Equation of normal
$y+3x\pm \sqrt { 3 } =0$

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

Length of latusrectum of the ellipse $\dfrac{x^{2}}{4}+\dfrac{y^{2}}{b^{2}}=1$, if the normal, at an end of latusrectum passes through one extremity of the minor axis, then equation of eccentricity of ellipse is

  1. $e^4+e^2-1=0$
  2. <font face="MathJax_Main">$e^3+e^2-1=0$</font>
  3. $e^4+e^2+1=0$

  4. <span class="MathJax_Preview"><span class="MathJax"><span class="math"><font face="MathJax_Main">none of these</font>
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a specific derivation problem involving the normal at the end of the latus rectum. The resulting condition for eccentricity e is e^4 + e^2 - 1 = 0.

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii
The domain of $f(x)=\dfrac 1{\sqrt {x-[x]}}$ is 
  1. $R$
  2. $Z$
  3. $R-Z$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The function is defined as 

$f(x)=\dfrac 1{\sqrt {x-[x]}}$

The function is not defined if

$\sqrt {x-[x]}=0$

$\implies x-[x]=0$

$\implies x=[x]$

This happens only in the case of Integers 

So The function is not defined at Integer Values

Hence , The Domain of function is $R-Z$

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The normal of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at a point $P(x _1,y _1)$ on  it, meets the x-axis in $G$. $PN$ is perpendicular to $OX$, where $O$ is origin. The value of $\frac{l(OG)}{l(ON)}$ is -

  1. $e$
  2. $e^2$
  3. $e^3$
  4. $e^2-1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For an ellipse x^2/a^2 + y^2/b^2 = 1, the foot of the normal on the x-axis gives an x-coordinate related to the center such that l(OG)/l(ON) simplifies to eccentricity squared, e^2, using standard ellipse coordinate geometry properties.

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The maximum number of normals that can be drawn from any point outside of an ellipse, in general, is 

  1. $2$
  2. $3$
  3. $1$
  4. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation of normals to an ellipse from a given point leads to a degree 4 polynomial in the eccentric angle parameters, meaning a maximum of 4 normals can be drawn from any point outside an ellipse.

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The line $y = mx - \displaystyle \frac{(a^2 - b^2)m }{\sqrt{a^2+ b^2 m^2}}$ is normal to the ellipse $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ for all values of $m$ belongs to:

  1. $(0, 1)$
  2. $(0, \infty)$
  3. $R$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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 $m  = \dfrac{a}{b} tan  \theta  \in  R.$

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

If the normal at an end of a latus-rectum of an ellipse $\displaystyle\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through one extremity of the minor axis, the eccentricity of the ellipse is given by:

  1. $e^2=5$
  2. $\displaystyle e^2=\frac{\sqrt{5}+1}{2}$
  3. $\displaystyle e=\frac{\sqrt{5}-1}{2}$
  4. $\displaystyle e^2=\frac{\sqrt{5}-1}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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,  $e^4+e^2-1=0$
$e^2=\cfrac{-1+\sqrt{5}}{2}, \cfrac{-1-\sqrt{5}}{2}$(not possible)

$\therefore e^2=\cfrac{-1+\sqrt{5}}{2}$