Tag: definitions related to parabola

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Questions Related to definitions related to parabola

Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola
Consider the conic $ x^{2}+4y-6x+k=0 $ & $\displaystyle L\Rightarrow y+1=0$ be its directrix
On the basis of above information answer the following question:

The focus of the parabola is

  1. $(-3, 3)$
  2. $(-3, -3)$
  3. $(3, -3)$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$x^2+4y-6x+k=0$
$\implies (x-3)^2=-4\left ( y-(\dfrac{k-9}{4}) \right )$
So, length of latus rectum $=4a=4$
$\therefore a=1$
Distance between vertex and directrix=a
$\therefore\left ( -1-(\dfrac{k-9}{4}) \right )=1$
$\therefore k=1$
So, co-ordinates of vertex are $\left(3,\dfrac{k-9}{4}\right)=(3,-2)$
Therefore, co-ordinates of focus are $(3,-2-a)=(3,-3)$
So, answer is option (C).
Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola
If the focus is $ \displaystyle (\alpha, \beta) $ & the directrix is $ \displaystyle ax+by+c=0 $ then the equation of conic whose eccentricity $=e $ is given by $ \displaystyle \left ( x-\alpha \right )^{2}+\left ( y-\beta \right )^{2}=e^{2}\frac{\left ( ax+by+c \right )^{2}}{a^{2}+b^{2}}$. If $e=1$ then conic is called parabola, for $ e < 1 $ (conic is an ellipse) and for $e > 1,$ conic is a hyperbola. 
Now consider the conic
$\displaystyle 169 \left \{\left (x-1  \right )^{2}+\left (y-3  \right )^{2}  \right \}=\left (5x-12y+17  \right )^{2} $ ......$(*)$
On the basis of above information answer the following question:
 The focus of the conic $(*)$ is
  1. $(-1, -3)$
  2. $(1,3)$
  3. $(5, -12)$
  4. $(-1, 3)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A conic is the locus of a point '$P$' which moves in such a way that its distances from a fixed point'$ S $' always bears a constant ratio to its distances from a fixed straight line. 
The fixed point '$S$' is called focus. 

The fixed straight line is called the directrix & the constant ratio is known as eccentricity denoted by $e$.
$\displaystyle \therefore e=\cfrac {PS}{PM}$
Now $\displaystyle 169\left { \left ( x-1 \right )^{2}+\left ( y-3 \right )^{2} \right }=\left ( 5x-12y+17 \right )^{2}$
$\displaystyle \Rightarrow \left { \left ( x-1 \right )^{2}+\left ( y-3 \right )^{2} \right }=\frac{\left ( 5x-12y+17 \right )^{2}}{5^{2}+(-12)^{2}}=\left ( \frac{5x-12y+17}{13} \right )^{2}$
$\displaystyle \therefore \left ( x-1 \right )^{2}+\left ( y-3 \right )^{2}=e^{2}\left ( \frac{5x-12+17}{13} \right )$ where $e=1$

Focus of conic $(* )$ is $(1, 3)$

Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

The vertex of the parabola $2((x-1)^2 + (y-2)^2) = (x + y + 3)^2$ is

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

Given parabola may be written as,$\displaystyle \sqrt {(x-1)^2+(y-2)^2}=\frac {|x+y+3|}{\sqrt 2}$
$\Rightarrow$  focus is $S=(1,2)$, diretrix is, $x + y + 3 = 0 ....(1)$
We know axis of the parabola passes through focus and perpendicular to the directrix.
Thus equation of axis is,$x -y + 1 = 0 .....(2)$
solving (1) and (2) we get the foot of directrix $P(-2,-1)$
So the vertex of the parabola will be mid point of PS
$\Rightarrow V = \displaystyle \left (-\frac {1}{2}, \frac {1}{2}\right )$

Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

If a point $\mathrm{P}$ moves such that the distance from the point $\mathrm{A} (1, 1)$ and the line $x+y+2=0$ are equal then the locus of $\mathrm{P}$ is equal to

  1. a straightline

  2. a parabola

  3. a pair of st. lines

  4. an ellipse
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let the coordinate at $P$ is $(h, k)$

$\therefore \ $ Distance com point $A(1, 1)$ is $AP=\sqrt {(n-1)^2+(k-1)^2}$

$\therefore \ $ Distance com line $x+y+2=0$ is $=\dfrac {h+k+2}{\sqrt {1^2 +1^2}}$

According to equation

$\sqrt {(h-1)^2 +(k-1)^2}=\dfrac {h+k+2}{\sqrt 2}$

Squaring both are

$2\left\{(h-1)^2 +(k-1)^2\right\}=(h+k+2)^2$

$\Rightarrow \ 2(h^2-2h+1+k^2-2k+1)=h^2+k^2+4+2hk+4h+4k$

$\Rightarrow \ 2h^2-4h+2k^2-4k+4=h^2+k^2+4+2hk+4h+4k$

$\Rightarrow \ h^2-hk+k^2=8h+4k$

$\Rightarrow \ (h-k)^2=8(h+k)$

Locus of $P$ is, $(x-y)^2=8(x+y)$ which is a parabola

Option $\to (B)$

Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

If the vertex of the conic $y^{2} - 4y = 4x - 4a$ always lies between the straight lines $x + y = 3$ and $2x + 2y - 1 = 0$ then

  1. $2 < a < 4$
  2. $-\dfrac {1}{2} < a < 2$
  3. $0 < a < 2$
  4. $-\dfrac {1}{2} < a < \dfrac {3}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Vertex of $y^{2} - 4y = 4x - 4a$ is $(a - 1, 2)$
So, $(a - 1 + 2 - 3)(2a - 2 + 4 - 1) < 0$
$(a - 2)(2a + 1) < 0$
$-\dfrac {1}{2} < a < 2$

Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

For the parabola $9x^{2} - 24xy + 16y^{2} - 20x - 15y - 60 = 0$ which of the following is/ are true.

  1. $focus = \left (-\dfrac {43}{25}, -\dfrac {129}{100}\right )$
  2. $focus = \left (\dfrac {43}{25}, \dfrac {129}{100}\right )$
  3. $directrix : 4x + 3y + \dfrac {53}{4} = 0$
  4. $directrix : 4x + 3y - \dfrac {53}{4} = 0$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

Two manually perpendicular tangent of the parabola ${ y }^{ 2 }=4ax$ meet the axis in ${P} _{1}$ and ${P} _{2}$. If $S$ is the focus of the parabola, then $\dfrac { 1 }{ \left( S{ P } _{ 1 } \right)  } +\dfrac { 1 }{ \left( S{ P } _{ 2 } \right)  } $ is equal to :-

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

${ y }^{ 2 }=4ax\quad ............(1)$

Let the two mutually perpendicular tangents have slopes ${ m } _{ 1 },{ m } _{ 1 }$ where ${ m } _{ 1 }=\cfrac { 1 }{ { m } _{ 2 } } $
And hence their questions be,
${ T } _{ 1 }:y=mx+\cfrac { a }{ m } ...........(2)$
${ T } _{ 2 }:y=\cfrac { 1 }{ m } x-am...........(3)$
Let ${ P } _{ 1 }\quad $and$\quad { P } _{ 2 }\quad $be$\quad ({ x } _{ 1 },0)&amp; ({ x } _{ 2 },0)$ these points must lie on ${ T } _{ 1 }$and${ T } _{ 2 }$
For ${ T } _{ 1 },\quad o=m({ x } _{ 1 })+\cfrac { a }{ m } $
       $=>{ x } _{ 1 }=-\cfrac { a }{ { m }^{ 2 } } $
For ${ T } _{ 2 },\quad o=-\cfrac { { x } _{ 2 } }{ m } -am$
       $=>{ x } _{ 2 }=-a{ m }^{ 2 }$
So, ${ P } _{ 1 }(\cfrac { -a }{ { m }^{ 2 } } ,0)\quad $and$\quad { P } _{ 2 }(-a{ m }^{ 2 },0)$
Focus $S=(a,o)$
Now $S{ P } _{ 1 }=\sqrt { (a+{ \cfrac { a }{ { m }^{ 2 } }  })^{ 2 }+0 } =(a+\cfrac { a }{ { m }^{ 2 } } )$
         $S{ P } _{ 2 }=\sqrt { (a+{ a{ m }^{ 2 } })^{ 2 }+0 } =(a+a{ m }^{ 2 })$
Now, $\cfrac { 1 }{ { SP } _{ 1 } } +\cfrac { 1 }{ { { SP } _{ 2 } } } =\cfrac { { m }^{ 2 } }{ a{ m }^{ 2 }+a } +\cfrac { 1 }{ a+{ am }^{ 2 } } $
                               $=\cfrac { { m }^{ 2 }+1 }{ a({ m }^{ 2 }+1) } $
                               $=\cfrac { 1 }{ a } $