Tag: two dimensional analytical geometry-ii

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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 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 : 

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

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

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

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 

  1. $5$
  2. $\dfrac{13}{3}$
  3. $\displaystyle \frac{15}{2}$
  4. $\displaystyle \frac{9}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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 $\displaystyle\frac{\pi}{4}.$ 

  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 } )} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The perpendicular distances from the center to the tangent and normal at eccentric angle pi/4 are p1 = ab/sqrt(b^2cos^2(pi/4) + a^2sin^2(pi/4)) and p2 = (a^2-b^2)sin(pi/4)cos(pi/4)/sqrt(a^2sin^2(pi/4) + b^2cos^2(pi/4)). The product of these distances for a rectangle formed by these perpendiculars is the area.

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

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 $\displaystyle \frac{a^{2}x}{x _1}-\frac{b^{2}y}{y _1}=a^{2}-b^2$

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 

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

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:

  1. $a+b$
  2. $a-b$
  3. $a^{2}+b^{2}$
  4. $a^{2}-b^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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) } } $ 
$\Rightarrow d=a-b$
Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The maximum distance of the normal to the ellipse $\displaystyle \frac{\mathrm{x}^{2}}{9}+\frac{\mathrm{y}^{2}}{4}=1$ from its centre is:

  1. $\displaystyle \frac{1}{2}$
  2. $2$
  3. $1$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Ellipse : $\cfrac { { x }^{ 2 } }{ 9 } +\cfrac { { y }^{ 2 } }{ 4 } =1$

Equation of the normal,
$\cfrac { { ax }^{  } }{ \cos  { \theta  }  } -\cfrac { { by }^{  } }{ \sin { \theta  } } ={ a }^{ 2 }-{ b }^{ 2 } \\ \therefore \cfrac { { 3x }^{  } }{ \cos  { \theta  }  } -\cfrac { { 2y }^{  } }{ \sin { \theta  } } =5$

Or, $\ { 3x }^{  }\sin { \theta  }-{ 2y }^{  }\cos  { \theta  } =5\cos  { \theta  } \sin { \theta  }$

Distance from origin d= $\cfrac { \left| 0+0-5\cos  { \theta  } \sin { \theta  } \right|  }{ \sqrt { 9{ \left( \cos  { \theta  }  \right)  }^{ 2 }+{ 4\left( \sin { \theta  } \right)  }^{ 2 } }  } $

Or, d=$\cfrac { 5 }{ \sqrt { 9{ \left( \csc { \theta  }  \right)  }^{ 2 }+{ 4 }{ \left( \sec { \theta  }  \right)  }^{ 2 } }  } $

To maximize d we need to minimize the denominator.
$E=9{ \left( \csc { \theta  }  \right)  }^{ 2 }+{ 4 }{ \left( \sec { \theta  }  \right)  }^{ 2 }\ then,\quad \\\cfrac { dE }{ d\theta  } =-18{ \left( \csc { \theta  }  \right)  }^{ 2 }\cot { \theta  } +8{ \left( \sec { \theta  }  \right)  }^{ 2 }\tan { \theta  } \ For\quad \\Minimizing,\quad \cfrac { dE }{ d\theta  } =0\\ \therefore -18{ \left( \csc { \theta  }  \right)  }^{ 2 }\cot { \theta  } +8{ \left( \sec { \theta  }  \right)  }^{ 2 }\tan { \theta  } =0 \\Or,18{ \left( \csc { \theta  }  \right)  }^{ 2 }\cot { \theta  } =8{ \left( \sec { \theta  }  \right)  }^{ 2 }\tan { \theta  } \\ Or,{ \left( \tan { \theta  }  \right)  }^{ 4 }=\cfrac { 9 }{ 4 } \\ Or,\quad \tan { \theta  } =\sqrt { \cfrac { 3 }{ 2 }  } \\ \therefore \csc { \theta  } =\sqrt { \cfrac { 5 }{ 3 }  } \quad \quad and\quad \quad \sec { \theta  } =\sqrt { \cfrac { 5 }{ 2 }  } $

 On putting the values in d we get,
$d=\cfrac { 5 }{ \sqrt { 15+10 }  } \\ Or,\quad d=1$
Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

lf the tangent drawn at a point $(t^{2},2t)$ on the parabola $y^{2}=4x$ is same as normal drawn at $(\sqrt{5}\cos\alpha, 2\sin\alpha)$ on the ellipse $\displaystyle \frac{x^{2}}{5}+\frac{y^{2}}{4}=1$, then which of following is not true?  

  1. $t=\displaystyle \pm\frac{1}{\sqrt{5}}$
  2. $\alpha=-\tan^{-1}2$
  3. $\alpha=\tan^{-1}2$
  4. $\alpha=\tan^{-1}4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The tangent to y^2=4x at (t^2, 2t) is ty = x + t^2. The normal to x^2/5 + y^2/4 = 1 at (sqrt(5)cos(alpha), 2sin(alpha)) is sqrt(5)xsec(alpha) - 2ycosec(alpha) = 1. Comparing coefficients leads to the condition for the lines to be identical, which excludes option D.

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

If the line $x\cos { \alpha  } +y\sin { \alpha  } =p$ be normal to the ellipse $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$, then

  1. ${ p }^{ 2 }\left( { a }^{ 2 }\cos ^{ 2 }{ \alpha } +{ b }^{ 2 }\sin ^{ 2 }{ \alpha } \right) ={ a }^{ 2 }-{ b }^{ 2 }$
  2. ${ p }^{ 2 }\left( { a }^{ 2 }\cos ^{ 2 }{ \alpha } +{ b }^{ 2 }\sin ^{ 2 }{ \alpha } \right) ={ \left( { a }^{ 2 }-{ b }^{ 2 } \right) }^{ 2 }$
  3. ${ p }^{ 2 }\left( { a }^{ 2 }\sec ^{ 2 }{ \alpha } +{ b }^{ 2 }\csc ^{ 2 }{ \alpha } \right) ={ a }^{ 2 }-{ b }^{ 2 }$
  4. ${ p }^{ 2 }\left( { a }^{ 2 }\sec ^{ 2 }{ \alpha } +{ b }^{ 2 }\csc ^{ 2 }{ \alpha } \right) ={ \left( { a }^{ 2 }-{ b }^{ 2 } \right) }^{ 2 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation of any normal to $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ is 
       $ax\sec { \phi  } -by\csc { \phi  } ={ a }^{ 2 }-{ b }^{ 2 }$              ......(i)
The straight line $x\cos { \alpha  } +y\sin { \alpha  } =p$ will be a normal to the ellipse $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$, if equation (i) and $x\cos { \alpha  } +y\sin { \alpha  } =p$ represent the same line.
$\therefore \dfrac { a\sec { \phi  }  }{ \cos { \alpha  }  } =\dfrac { -b\csc { \phi  }  }{ \sin { \alpha  }  } =\dfrac { { a }^{ 2 }-{ b }^{ 2 } }{ p } $
$\Rightarrow \cos { \phi  } =\dfrac { ap }{ \left( { a }^{ 2 }-{ b }^{ 2 } \right) \cos { \alpha  }  } $
$\sin { \phi  } =\dfrac { -bp }{ \left( { a }^{ 2 }-{ b }^{ 2 } \right) \sin { \alpha  }  } $
$\because \sin ^{ 2 }{ \phi  } +\cos ^{ 2 }{ \phi  } =1$
$\Rightarrow \dfrac { { b }^{ 2 }{ p }^{ 2 } }{ { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 }\sin ^{ 2 }{ \alpha  }  } +\dfrac { { a }^{ 2 }{ p }^{ 2 } }{ { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 }\cos ^{ 2 }{ \alpha  }  } =1$
$\Rightarrow { p }^{ 2 }\left( { b }^{ 2 }\csc ^{ 2 }{ \alpha  } +{ a }^{ 2 }\sec ^{ 2 }{ \alpha  }  \right) ={ \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 }$

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

If the line $x \cos a + y \sin a = p$ be normal to the ellipse $\dfrac{x^2}{a^2}$ $+\dfrac{y^2}{b^2}$ = 1 then

  1. $p^2(a^2\cos^2a+b^2\sin^2a)=a^2-b^2$
  2. $p^2(a^2\cos^2a+b^2\sin^2a)=(a^2-b^2)^2$
  3. $p^2(a^2\sec^2a+b^2\csc^2a)=(a^2-b^2)$
  4. $p^2(a^2\sec^2a+b^2\csc^2a)=(a^2-b^2)^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A line $y=mx+c$ is normal to ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ if $c^2=m^2\dfrac{(a^2-b^2)^2}{a^2+m^2b^2}$
Given equation $x\cos a+y\sin a=p\Rightarrow y=-(\dfrac{\cos a}{\sin a})x+\dfrac{p}{\sin a}$
Here $m=-\cot a, c=\dfrac{p}{\sin a}$
Substituting in the formulae, we get
$\dfrac{p^2}{\sin ^2a}=\dfrac{\cos ^2a}{sin^2a}\times \dfrac{(a^2-b^2)^2}{a^2+(\cot^2a) b^2}$
After simplification, we get
$p^2\dfrac{(a^2\sin^2a+b^2\cos^2a)}{\sin^2a\cos^2a}=(a^2-b^2)^2$
$p^2(a^2\sec^2a+b^2\csc^2a)=(a^2-b^2)^2$

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

If the normal at the point $P(\theta)$ to the ellipse $\dfrac {x^{2}}{14} + \dfrac {y^{2}}{5} = 1$ intersects it again at the point $Q(2\theta)$, then $\cos \theta$ is equal to

  1. $2/3$
  2. $-2/3$
  3. $3/4$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Normal at the point P(theta) to the ellipse $\dfrac{x²}{14} + \dfrac{y²}{5 }= 1$ intersects it again at the point $Q(2\  \theta). $

we know, standard equation of ellipse is 

$\dfrac{x²}{a²} + \dfrac{y²}{b²} = 1 $ compare it with given equation

so, $ a² = 14$ then, $a = √14 $

$b² = 5$ then, $b = √5 $

now equation of normal passing through point $P(\theta)$ is given by, 

$\dfrac{ax}{cos \theta} - \dfrac{by}{sin \theta} = a² - b². $

or, $\dfrac{\sqrt14x}{cos \theta} -\dfrac{ \sqrt5y}{sin \theta} = 14 - 5 = 9$ ....(1) 

it again meets the curve at the point $Q(2\theta) $

so, $Q(2\theta) = (√14cos2\theta, √5sin2\theta) $

now, put it in equation (1), 

or, $\dfrac{14cos2\theta}{\cos \theta} - \dfrac{5sin2\theta}{\sin\theta} = 9$ 

or, ${14(2cos² \theta - 1)}{\cos \theta} - \dfrac{10sin\theta cos \theta}{\sin \theta} = 9$

or, $28cos \theta - 14sec \theta - 10cos \theta = 9$

or, $18cos \theta - \dfrac{14}{cos \theta} = 9$

or, $18cos²\theta - 14 - 9cos \theta = 0$

or, $18cos²\theta -21cos \theta + 12cos\theta - 14 = 0$

or, $3cos \theta(6cos \theta - 7) + 2(cos \theta - 7) = 0$

or, $(3cos \theta + 2)(6cos \theta - 7) = 0$

or, $cos \theta = \dfrac{-2}{3} $