Tag: normal to an ellipse

Questions Related to normal to an ellipse

Find where the line $\displaystyle 2x+y=3$ cuts the curve $\displaystyle 4x^{2}+y^{2}=5.$ Obtain the equations of the normals at the points of intersection and determine the co-ordinates of the point where these normals cut each other.

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

  1. $\displaystyle \left ( -1, \frac{1}{2} \right )$

  2. $\displaystyle \left ( 1, \frac{1}{2} \right )$

  3. $\displaystyle \left ( -1, \frac{-1}{2} \right )$

  4. $\displaystyle \left ( 1, \frac{-1}{2} \right )$

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Correct Option: A
Explanation:

$\displaystyle P\left ( \frac{1}{2}+, 2 \right ), Q(1, 1)$ 
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 $\displaystyle \left ( -1, \frac{1}{2} \right )$

If $y=mx+7\sqrt{3}$ is normal to $\dfrac{x^2}{18}-\dfrac{y^2}{24}=1$ then the value of m can be?

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

  1. $\dfrac{2}{\sqrt{5}}$

  2. $\dfrac{4}{\sqrt{5}}$

  3. $\dfrac{1}{\sqrt{5}}$

  4. $\dfrac{2}{\sqrt{3}}$

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Correct Option: A
Explanation:

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

The normal at a point $P$ on the ellipse $x^{2}+4y^{2}=16$ meets the x-axis at $Q.$ If $M$ is the mid point of the line segment $PQ$, then locus of $M$ intersects the latus rectums of the given ellipse at the points.

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

  1. $\displaystyle \left ( \pm \frac{3\sqrt{5}}{7}, \pm \frac{2}{7} \right )$

  2. $\displaystyle \left ( \pm \frac{3\sqrt{5}}{2}, \pm \frac{\sqrt{19}}{4} \right )$

  3. $\displaystyle \left ( \pm 2\sqrt{3}, \pm \frac{1}{7} \right )$

  4. $\displaystyle \left ( \pm 2\sqrt{3}, \pm \frac{4\sqrt{3}}{7} \right )$

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Correct Option: C
Explanation:

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

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

Latus rectum of the given ellipse is
$x=\pm ae=\pm \sqrt{16-4}=\pm 2\sqrt{3}$
So locus of $M$ meets the latus rectum at points for which
$\displaystyle y^{2}=1-\frac{12\times 4}{49}=\frac{1}{49}$   $\Rightarrow $   $\displaystyle y=\pm \frac{1}{7}$
Hence, the required point is $\displaystyle \left ( \pm 2\sqrt{3}, \pm \frac{1}{7} \right )$.

The eccentric angle of the point where the line, $5x\, -\, 3y\, =\, 8\sqrt{2}$ is a normal to the ellipse $\displaystyle\frac{x^2}{25}\, +\, \frac{y^2}{9}\,=\,1$ is

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

  1. $\displaystyle\frac{3\pi}{4}$

  2. $\displaystyle\frac{\pi}{4}$

  3. $\displaystyle\frac{\pi}{6}$

  4. $tan^{-1}\,2$

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Correct Option: B
Explanation:

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,ellipse equation as $\dfrac{x^2}{5^2} + \dfrac{y^2}{3^2} = 1$

$\Rightarrow$ Length of major axis, $a=5$ and length of minor axis, $b=3$.

$\therefore$The required equation of normal is $ 5x\sec\theta-3ycosec\theta=5^2-3^2$

$\Rightarrow 5x\sec\theta-3ycosec\theta=16 \dots (1)$

Given normal equation $5x-3y=8\sqrt2$

Multiplying both sides with $\sqrt2$

$\Rightarrow 5\sqrt2 x-3\sqrt2 y=16\dots (2)$

Comparing equation $(1)$ and $(2)$

$\Rightarrow \sec\theta=\sqrt2$

$\Rightarrow \cos\theta=\dfrac{1}{\sqrt2}$

$\Rightarrow \theta =\dfrac{\pi}{4}$

On the ellipse $\displaystyle \frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 9 } =1$, one of the points at which the normals are parallel to the line $2x-y=1$ is

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

  1. $\displaystyle \left( \frac { 9 }{ \sqrt { 10 }  } ,\frac { 2 }{ \sqrt { 10 }  }  \right) $

  2. $\displaystyle \left( -\frac { 9 }{ \sqrt { 10 }  } ,\frac { 2 }{ \sqrt { 10 }  }  \right) $

  3. $\displaystyle \left( \frac { 2 }{ \sqrt { 10 }  } ,\frac { 9 }{ \sqrt { 10 }  }  \right) $

  4. None of these

Show answer
Correct Option: C
Explanation:

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.

The equation of the normal to the ellipse $\displaystyle\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}\,=\,1$ at the positive end of latus rectum is : 

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

  1. $x\,+\,ey\,+\,e^2a\,=\,0$

  2. $x\,-\,ey\,-\,e^3a\,=\,0$

  3. $x\,-\,ey\,-\,e^2a\,=\,0$

  4. none of these

Show answer
Correct Option: B
Explanation:

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


Option [B]

Area of the triangle formed by the ${x}$ axis, the tangent and normal at $(3,2)$ to the ellipse $\displaystyle \frac{x^{2}}{18}+\frac{y^{2}}{8}=1$ is 

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

  1. $5$

  2. $\dfrac{13}{3}$

  3. $\displaystyle \frac{15}{2}$

  4. $\displaystyle \frac{9}{2}$

Show answer
Correct Option: B
Explanation:

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

Find the area of the rectangle formed by the perpendiculars from the center of the ellipse $\displaystyle \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ to the tangent and normal at a point whose eccentric angle is $\displaystyle\frac{\pi}{4}.$ 

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

  1. $\displaystyle \frac { \left( { a }^{ 2 }-{ b }^{ 2 } \right) ab }{ { a }^{ 2 }+{ b }^{ 2 } } $

  2. $\displaystyle \frac { \left( { a }^{ 2 }+{ b }^{ 2 } \right) ab }{ { a }^{ 2 }-{ b }^{ 2 } } $

  3. $\displaystyle \frac { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }{ab( { a }^{ 2 }+{ b }^{ 2 } )} $

  4. $\displaystyle \frac { \left( { a }^{ 2 }+{ b }^{ 2 } \right)  }{ab( { a }^{ 2 }-{ b }^{ 2 } )} $

Show answer
Correct Option: A

Assertion (A): Equation of the normal to the ellipse $\displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{9}=1$ at $P(\displaystyle \frac{\pi}{4})$ is $5x-3y-8\sqrt{2}=0$
Reason (R): Equation of the normal to the ellipse $\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at $P(x _{1},y _{1})$ is $\displaystyle \frac{a^{2}x}{x _1}-\frac{b^{2}y}{y _1}=a^{2}-b^2$

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

  1. Both A and R are true but R is not the correct explanation of A

  2. Both A and R are true and R is the correct explanation of A

  3. A is true but R is false

  4. A is false but R is True

Show answer
Correct Option: B
Explanation:

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 maximum distance of any normal to the ellipse $\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ from the centre is:

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

  1. $a+b$

  2. $a-b$

  3. $a^{2}+b^{2}$

  4. $a^{2}-b^{2}$

Show answer
Correct Option: B
Explanation:
Equation of any normal to the ellipse is:
$ \cfrac { ax }{ \cos { \theta  } } -\cfrac { by }{ \sin { \theta  } } =ae$ 
Distance from the center is: $ d=\cfrac { ae }{ \sqrt { \cfrac { { a }^{ 2 } }{ { (\cos { \theta  }) }^{ 2 } } +\cfrac { { b }^{ 2 } }{ { (\sin { \theta  }) }^{ 2 } }  }  }$        -------(1) 
$a,e,b$ are fixed for a given ellipse. So, to maximize $d$ we need to minimize the denominator
$ \therefore E={ a }^{ 2 }({ \sec { \theta  }) }^{ 2 }+{ b }^{ 2 }{ \left( co\sec { \theta  } \right)  }^{ 2 }$
$ \cfrac { DE }{ D\theta  } =2{ a }^{ 2 }({ \sec { \theta  }) }^{ 3 }\tan \theta -{ 2b }^{ 2 }{ \left( co\sec { \theta  } \right)  }^{ 3 }\cot \theta =0$
$ \Rightarrow 2{ a }^{ 2 }({ \sec { \theta  }) }^{ 3 }\tan \theta ={ 2b }^{ 2 }{ \left( co\sec { \theta  } \right)  }^{ 3 }\cot \theta $ 
$\Rightarrow { (\tan \theta ) }^{ 4 }=\cfrac { { b }^{ 2 } }{ { a }^{ 2 } } $ 
$\Rightarrow \tan \theta =\sqrt { \cfrac { b }{ a } } $ 
$\therefore  \sin \theta =\sqrt { \cfrac { b }{ a+b }  }$ and 
$\cos \theta =\sqrt { \cfrac { a }{ a+b }  } $ 
Putting these values in equation 1 we get:
$ d=\cfrac { ae }{ \sqrt { \cfrac { { a }^{ 2 }(a+b) }{ a } +\cfrac { { b }^{ 2 }(a+b) }{ b }  }  } $ 
$\Rightarrow d=\cfrac { ae }{ { (a+b) } } $ 
$\Rightarrow d=\cfrac { ae(a-b) }{ { (a+b)(a-b) } } $ 
$\Rightarrow d=a-b$