Tag: ellipse

Quizzes (5) Questions (224)

Questions Related to ellipse

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

The equation of normal at the point $(0, 3)$ of the ellipse $9x^2 + 5y^2 = 45$ is

  1. $y- 3 = 0$
  2. $y + 3 = 0$
  3. $x$-axis
  4. $y$-axis
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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


Differentiating with respect to $x$, we get,

$18x+10y\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}=-\dfrac{18x}{10y}$

Therefore, Slope of tangent at $(0,3)=\dfrac{-0}{30}=0$

Slope of normal = $\dfrac{1}{0}$

The equation of normal is :
$y-3=\dfrac{1}{0}(x-0)$
$x-0=0$
$x=0$

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

Find the equation of the normal to the ellipse $9x^2 + 16y^2 = 288$ at the point $(4, 3).$

  1. $4x-3y=7$
  2. $3x-4y=7$
  3. $4x+3y=7$
  4. $3x+4y=7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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,
$\displaystyle y-3=\frac{3\times 32}{4\times 18}(x-4)\Rightarrow y-3=\frac{4}{3}(x-4)\Rightarrow 4x-3y=7$

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 $R-Z$