Tag: normal to an ellipse

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Questions Related to normal to an ellipse

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

The line $5x - 3y = 8\sqrt{2}$ is a normal to the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$. If $\theta$ be the eccentric angle of the foot of this normal , then '$\theta$' is equal to

  1. $\dfrac{\pi}{6}$
  2. $\dfrac{\pi}{3}$
  3. $\dfrac{\pi}{4}$
  4. $\dfrac{\pi}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
From the equation of ellipse given, we have; $a=5,b=3$

Therefore any point on the ellipse $\left( 5cos\theta , 3sin\theta  \right) .$

Normal at this point to the given ellipse is 
$ 5x sec\theta -3y cosec\theta =25-9=16------(1)$

 Also the equation of given normal is $ 5x-3y=8\sqrt { 2 } -----(2)$

 Also the equation of the normal $(1)$ and $(2)$,

$ \Longrightarrow \dfrac { 5sec\theta  }{ 5 } =\dfrac { 3cosec\theta  }{ 3 } =\dfrac { 16 }{ 8\sqrt { 2 }  }$

$ \Longrightarrow sin\theta =cos\theta =\frac { 1 }{ \sqrt { 2 }  }$

$\Longrightarrow \theta =\frac { \Pi  }{ 4 } $

Option (c) is correct.
Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

Let $L$ be an end of the latus rectum of $y^2 = 4x$. The normal at $L$ meets the curve again at $M$. The normal at $M$ meets the curve again at $N$. The area of $\Delta LMN$ is

  1. $\dfrac{1280}{9} sq.$ units
  2. $\dfrac{640}{9} sq.$ units
  3. $\dfrac{320}{9} sq.$ units
  4. $\dfrac{160}{9} sq.$ units
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the parabola y^2=4x, the latus rectum end L is (1, 2). The normal at (1, 2) is y+2x=4, which meets the parabola at M(9, -6). The normal at M is y+x/2 = -3, which meets the parabola at N(1/4, -1). The area of triangle LMN with vertices (1, 2), (9, -6), and (0.25, -1) is 1280/9.

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

The line $2x+y =3$ cuts the ellipse $4x^2+y^2 =5$ at P and Q . If $\theta$ be the angle between the normals  at these point then $tan \theta$ =

  1. $1/2$
  2. $3/4$
  3. $3/5$
  4. $5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The line 2x+y=3 intersects the ellipse 4x^2+y^2=5 at P(1, 1) and Q(1/2, 2). The slopes of the normals at these points are calculated using the derivative dy/dx = -4x/y. The slope of the normal at P(1, 1) is 1/4 and at Q(1/2, 2) is 1. The angle theta between them satisfies tan(theta) = |(1 - 1/4) / (1 + 1*1/4)| = (3/4) / (5/4) = 3/5.

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 $\displaystyle x^{2} + 4y^{2} = 16$ at the end of the latus rectum in the first quadrant is

  1. $\displaystyle 2x + \sqrt{3} \left ( y + 3 \right ) = 0$
  2. $\displaystyle 2x = \sqrt{3} \left ( y+ 3 \right )$
  3. $\displaystyle \sqrt{3} x = 2 \left ( y + 3 \right )$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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 $(ae, \cfrac{b^2}{a})\equiv (2\sqrt{3},1)$

Then the point form equation of normal at point $(x _1,y _1)$ is $y-y _1=\displaystyle\frac {y _1a^2}{x _1b^2}(x-x _1), x _1\neq 0$
Thus equation of normal at this point is given by,
$\cfrac{16x}{2\sqrt{3}}-\cfrac{4y}{1}=12$
$\Rightarrow 2x=\sqrt{3}(y+3)$

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

If the tangent drawn at a point $\left( { t }^{ 2 },2t \right) $ on the parabola ${ y }^{ 2 }=4x$ is same as normal drawn at $\left( \sqrt { 5 } \cos { \alpha  } ,2\sin { \alpha  }  \right) $ on the ellipse $\displaystyle \frac { { x }^{ 2 } }{ 5 } +\frac { { y }^{ 2 } }{ 4 } =1$, then which of following is true.

  1. $\displaystyle t=\pm \frac { 1 }{ \sqrt { 5 } } $
  2. $\alpha =-\tan ^{ -1 }{ 2 } $
  3. $\alpha =\tan ^{ -1 }{ 2 } $

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

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


Equation of normal to ellipse $\displaystyle \frac { { x }^{ 2 } }{ 5 } +\frac { { y }^{ 2 } }{ 4 } =1$ at $\left( \sqrt { 5 } \cos { \alpha  } ,2\sin { \alpha  }  \right) $ is $\sqrt { 5 } \sec { \alpha x-2y\csc { \alpha  } =1 } $    ....(2)

Given (1) $=$ (2)

$\displaystyle \Rightarrow \sqrt { 5 } \sec { \alpha  } =\frac { 2\csc { \alpha  }  }{ t } =-\frac { 1 }{ { t }^{ 2 } } \Rightarrow \cos { \alpha  } =-\sqrt { 5 } { t }^{ 2 }$ and $\sin { \alpha  } =-2t$

$\displaystyle \Rightarrow \cos ^{ 2 }{ \alpha  } +\sin ^{ 2 }{ \alpha  } =5{ t }^{ 4 }+4{ t }^{ 2 }=1\Rightarrow { t }^{ 2 }=\frac { 1 }{ 5 } $   

$\therefore$ (A) is true
and $\displaystyle \frac { \sin { \alpha  }  }{ \cos { \alpha  }  } =-\frac { 2t }{ -\sqrt { 5 } { t }^{ 2 } } =\frac { 2 }{ \sqrt { 5 }  } \times \frac { 1 }{ t } =\frac { 2 }{ \sqrt { 5 }  } \times \left( \pm 5 \right) $

$\therefore \tan { \alpha  } =\pm 2$

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

The number of distinct normal lines from the exterior point $\displaystyle \left ( 0, : c \right ), : c > b$ , to the ellipse $\displaystyle \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is

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

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 $P(0,c)$
$\Rightarrow c = -\cfrac{(a^2-b^2)m}{\sqrt{a^2+b^2m^2}}$
$\Rightarrow c^2=\cfrac{(a^2-b^2)^2m^2}{a^2+b^2m^2}$
$\Rightarrow m^2=\cfrac{c^2a^2}{(a^2-b^2)^2-c^2b^2}$


Clearly this is a polynomial of degree two so maximum number of normal that
can be drawn from point $P (0,c)$  to the ellipse is $2$, with slopes $+\infty$ and $-\infty$ corresponding to  two roots of $m.$
But the question asked for distinct normal lines, therefore answer is $1$.

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

Equation of the normal to the ellipse  $4 ( x - 1 ) ^ { 2 } + 9 ( y - 2 ) ^ { 2 } = 36 ,$  which is parallel to the line  $3 x - y = 1 ,$  is

  1. $3 x - y = \sqrt { 5 }$
  2. $3 x - y = \sqrt { 5 } - 3$
  3. $3 x - y = \sqrt { 5 } + 2$
  4. $3 x - y = \sqrt { 5 } ( \sqrt { 5 } + 1 )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The ellipse is (x-1)^2/9 + (y-2)^2/4 = 1. The normal parallel to 3x-y=1 has slope -1/3. Using the normal form y-k = m(x-h) +/- m(a^2-b^2)/sqrt(a^2+b^2m^2), where m is the slope of the normal, we substitute the values to find the equation.

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

The number of normals that can be drawn to the curve $\displaystyle 4x^{2} + 9y^{2} = 36$ from an external point, in general, is

  1. $1$
  2. $3$
  3. $4$

  4. infinite
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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 $m.$

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

If  the equation of normal to the ellipse $\displaystyle  4x^{2}+9y^{2}=36$ at the point $(3, -2)$ is $ px+qy=r$. Find the value of $p+q+r.$

  1. $8$
  2. $9$
  3. $10$
  4. $12$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

$\Rightarrow \dfrac{9x}{3}-\dfrac{4y}{-2}=5$
$\Rightarrow 3x+2y=5$
Therefore, $p+q+r=10$

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

Find the condition that the line  $lx+my=n$ be a normal for ellipse

  1. $\displaystyle \frac{a^{2}}{l^{2}}-\frac{b^{2}}{m^{2}}=\frac{\left ( a^{2}+b^{2} \right )^{2}}{2n^{2}}$
  2. $\displaystyle \frac{a^{2}}{l^{2}}-\frac{b^{2}}{m^{2}}=\frac{\left ( a^{2}+b^{2} \right )^{2}}{n^{2}}$
  3. $\displaystyle \frac{a^{2}}{l^{2}}+\frac{b^{2}}{m^{2}}=\frac{\left ( a^{2}-b^{2} \right )^{2}}{n^{2}}$
  4. $\displaystyle \frac{a^{2}}{l^{2}}+\frac{b^{2}}{m^{2}}=\frac{\left ( a^{2}-b^{2} \right )^{2}}{2n^{2}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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.