Tag: ellipse

Questions Related to ellipse

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

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

  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

Show answer
Correct Option: B
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,
$\cfrac{16x}{2\sqrt{3}}-\cfrac{4y}{1}=12$
$\Rightarrow 2x=\sqrt{3}(y+3)$

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.

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

  1. $\displaystyle t=\pm \frac { 1 }{ \sqrt { 5 }  } $

  2. $\alpha =-\tan ^{ -1 }{ 2 } $

  3. $\alpha =\tan ^{ -1 }{ 2 } $

  4. None of these

Show answer
Correct Option: A,B,C
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) $

$\therefore \tan { \alpha  } =\pm 2$

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

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

  1. $3$

  2. $4$

  3. $2$

  4. $1$

Show answer
Correct Option: D
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 $1$.

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

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

  1. $3 x - y = \sqrt { 5 }$

  2. $3 x - y = \sqrt { 5 } - 3$

  3. $3 x - y = \sqrt { 5 } + 2$

  4. $3 x - y = \sqrt { 5 } ( \sqrt { 5 } + 1 )$

Show answer
Correct Option: A

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

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

  1. $1$

  2. $3$

  3. $4$

  4. infinite

Show answer
Correct Option: C
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 $m.$

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 $p+q+r.$

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

  1. $8$

  2. $9$

  3. $10$

  4. $12$

Show answer
Correct Option: C
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, $p+q+r=10$

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

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

  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. $\displaystyle \frac{a^{2}}{l^{2}}+\frac{b^{2}}{m^{2}}=\frac{\left ( a^{2}-b^{2} \right )^{2}}{2n^{2}}$

Show answer
Correct Option: C
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.

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

Show answer
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. $\dfrac{2}{\sqrt{3}}$

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

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