Tag: two dimensional analytical geometry-ii

Quizzes (14) Questions (88)

Questions Related to two dimensional analytical geometry-ii

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.

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

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.

  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. $\displaystyle \left ( 1, \frac{-1}{2} \right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\displaystyle \left ( -1, \frac{1}{2} \right )$

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

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?

  1. $\dfrac{2}{\sqrt{5}}$
  2. $\dfrac{4}{\sqrt{5}}$
  3. $\dfrac{1}{\sqrt{5}}$
  4. $\dfrac{2}{\sqrt{3}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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.

  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. $\displaystyle \left ( \pm 2\sqrt{3}, \pm \frac{4\sqrt{3}}{7} \right )$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $\displaystyle \left ( \pm 2\sqrt{3}, \pm \frac{1}{7} \right )$.

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

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

  1. $\displaystyle\frac{3\pi}{4}$
  2. $\displaystyle\frac{\pi}{4}$
  3. $\displaystyle\frac{\pi}{6}$
  4. $tan^{-1}\,2$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $(2)$

$\Rightarrow \sec\theta=\sqrt2$

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

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

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

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

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.