Tag: tracing of the parabola

Questions Related to tracing of the parabola

For parabola $x^{ 2 } + y^{ 2 } + 2xy 6x 2y + 3 = 0$ the focus is

  1. $\left( 1, -1\right)$

  2. $\left( -1, 1\right)$

  3. $\left( 3, 1\right)$

  4. $None of these$


Correct Option: A
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


Correct Option: C
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).
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. $(-1, 3)$


Correct Option: B
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 $(1, 3)$

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


Correct Option: B
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
$\Rightarrow V = \displaystyle \left (-\frac {1}{2}, \frac {1}{2}\right )$

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


Correct Option: B
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 $\to (B)$

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. $-\dfrac {1}{2} < a < \dfrac {3}{2}$


Correct Option: B
Explanation:

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

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. $directrix : 4x + 3y - \dfrac {53}{4} = 0$


Correct Option: B,C

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 } $


Correct Option: C
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 } $

A parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point in the plane.

  1. True

  2. False


Correct Option: A
Explanation:

Parabola is a set of all points in a plane that are equidistant from a fixed line (called directrix) and form a fixed point called focus.

The equation of a parabola in its standard form is $x^3=4ay.$

  1. True

  2. False


Correct Option: B
Explanation:

The given statement is wrong, because standard form the equation of a parabola is $y = ax^2+bx+c$, there is no cube root in equation of a parabola.