Tag: tracing of the parabola

Questions Related to tracing of the parabola

The solution of \$frac{{dy}}{{dx}} = \frac{{ax + h}}{{by + k}}$ represents a parabola

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

  1. a=0,b=0

  2. a=1, b=2

  3. a=0, b=0

  4. a=2, b=1

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Correct Option: A

The equation $y^2+3 =2( 2x +y)$ represents a parabola with vertex at 

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

  1. $\left(\dfrac{1}{2}, 1\right) $ and axis parallel to $y$-axis

  2. $\left(\dfrac{1}{2}, 1\right) $ and axis parallel to $ x$-axis

  3. $\left(\dfrac{1}{2}, 1\right) $ and focus at $\left(\dfrac{3}{2}, 1\right)$

  4. $\left(1, \dfrac{1}{2},\right) $ and focus at $\left(\dfrac{3}{2}, 1\right)$

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Correct Option: B,C
Explanation:

$y^2+3=2(2x+y)$ represents parabola.


$y^2+3=4x+2y$


$y^2-2y+3=4x$

$y^2-2y+1+3=4x+1$

$(y-1)^2=4x-2$

$(y-1)^2=4(x-\dfrac{1}{2})$

So, the vertex of parabola$=\left(\dfrac{1}{2},1\right)$ and axis is parallel to x axis.

$a=1$

Focus$=\left(\dfrac{1}{2}+1,1\right)$

               $=\left(\dfrac{3}{2},1\right)$

If the equation of parabola is ${x}^{2}=-9y$, then the equation of the directrix and the length of latus rectum are 

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

  1. $y=-\dfrac {9}{4}, 8$

  2. $x=\dfrac {-9}{4}, 9$

  3. $y=\dfrac {9}{4}, 9$

  4. $None\ of\ these$

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Correct Option: C
Explanation:
The given equation is of the form ${x}^{2} = - 4ay$, where a is positive.
Therefore, the focus is on the y-axis in the negative direction and parabola opens downwards.
Therefore,
Given equation of parabola-
${x}^{2} = -9 y ..... \left( 1 \right)$
Standard equation of parabola-
${x}^{2} = -4ay ..... \left( 2 \right)$
Comparing ${eq}^{n} \left( 1 \right) \& \left( 2 \right)$, we have
$a = \cfrac{9}{4}$
As we know that, for parabola in the form ${x}^{2} = -4ay$, equation of directrix for parabola is-
$y = a$
$\therefore$ For ${x}^{2} = -9y$, equation of directrix is-
$y = \cfrac{9}{4}$
Length of latus rectum, $l = 4a$
As $a = \cfrac{9}{4}$,
$\therefore \; l = 4 \times \cfrac{9}{4} = 9$
Hence, the equation of directrix will be $y = \cfrac{9}{4}$ and the lngth of latus rectum will be 9.

If $\displaystyle \left ( 2,0 \right )$ is the vertex and $y -$ axis the directrix of a parabola,find the coordinates of focus. 

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

  1. Focus is $\displaystyle \left ( 2,0 \right )$

  2. Focus is $\displaystyle \left ( 4,0 \right )$

  3. Focus is $\displaystyle \left ( 8,0 \right )$

  4. Focus is $\displaystyle \left ( -4,0 \right )$

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Correct Option: B
Explanation:

Distance of vertex from directrix $y -$ axis is $2$ and we know it
is half the distance of focus from directrix.
$\displaystyle \therefore $ Focus is $\displaystyle \left ( 4,0 \right ).$

The focal distance of a point $P$ on the parabola $y^2=12x$ if the ordinate of $P$ is $6$, is

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

  1. $12$

  2. $6$

  3. $3$

  4. $9$

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Correct Option: B
Explanation:

Given parabola is $y^2=12x$    ....$(i)$
Here $a=3$
For point $P(x, y), y=6$
This point lie on the parabola 
$\therefore (6)^2=12x\Rightarrow x=3$
Now, focal distance of point $P$ is $x+a=6$

The equation of the conic with focus $\displaystyle S \left( \frac{3}{2}, 0 \right) $ and the directrix 2x + 3 = 0 having eccentricity 1, is

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

  1. $y^2 = 4x$

  2. $y^2 = 5x$

  3. $y^2 = 6x$

  4. $y^2 = 8x$

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Correct Option: C

The locus of the points which are equidistant from $(-a, 0)$ and $x=a$ is

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

  1. $y^2=4ax$

  2. $y^2+4ax=0$

  3. $x^2+4ay=0$

  4. $x^2-4ay=0$

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Correct Option: B
Explanation:
Let $(h,k)$ be any point on the curve.
Distance of this point from $(-a,0)=\sqrt{(h-(-a))^2+(k-0)^2}$
Distance of point $(h,k)$ from the line $x-a=0$ is $\dfrac{h-a}{1}=h-a$
Point $(h,k)$ is equidistant from $(-a,0)$ and the line $x-a=0$
$\implies  \sqrt{(h-(-a))^2+(k-0)^2}=h-a$

$\implies \sqrt{(h+a)^2+(k-0)^2}=h-a$

Squaring the above equation, we get
$\implies (h+a)^2+(k-0)^2=(h-a)^2$
$\implies h^2+a^2+2ah+k^2=h^2+a^2-2ah$

$\implies k^2=-4ah$
Subtitute $k=y$ and $h=x$, we get

$\implies y^2=-4ax$
$\implies y^2+4ax=0$
So, the answer is option (B)


Find the equation of the parabola whose focus is $S(3,5)$ and vertex is $A(1,3)$.

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

  1. $\begin{array}{}\ \Rightarrow \left|  \right| =  {\left( {x + y} \right)^2} = 2\left[ {{{\left( {x - 3} \right)}^2} + {{\left( {y - 5} \right)}^2}} \right]\end{array}$

  2. $\begin{array}{}\ \Rightarrow \left|  \right| =  {\left( {x + y} \right)^2} = 2\left[ {{{\left( {x - 6} \right)}^2} + {{\left( {y - 6} \right)}^2}} \right]\end{array}$

  3. $\begin{array}{}\ \Rightarrow \left|  \right| =  {\left( {x + y} \right)^2} = 2\left[ {{{\left( {x - 11} \right)}^2} + {{\left( {y - 11} \right)}^2}} \right]\end{array}$

  4. $\begin{array}{}\ \Rightarrow \left|  \right| =  {\left( {x + y} \right)^2} = 2\left[ {{{\left( {x - 7} \right)}^2} + {{\left( {y - 7} \right)}^2}} \right]\end{array}$

Show answer
Correct Option: A
Explanation:

Slope of axis$=\frac{{5 - 3}}{{3 - 1}} = \frac{2}{2} = 1$
Slope of directrix$=-1$
equation of tangent at vertex $A$
$\begin{array}{l}Pt(1,3),m =  - 1\ \Rightarrow y - 3 =  - 1\left( {x - 1} \right)\ \Rightarrow x + y - 4 = 0\end{array}$
equation of directrix
$\begin{array}{l}x + y = \lambda \a = SA\ = \sqrt {4 + 4}  = 2\sqrt 2 \end{array}$
$A$ is midpoint of $PS$
$\begin{array}{l}\frac{{n + 3}}{2} = 1,\frac{{k + 5}}{2} = 3\ \Rightarrow n =  - 1,k = 1\end{array}$
$(-1,1)$ lies on directrix
$-1+1=\lambda=0$
equation of diectrix: $L:y+x=0$
$\begin{array}{l}QO = QS\ \Rightarrow \left| {\frac{{l + m}}{{\sqrt 2 }}} \right| = \sqrt {{{\left( {l - 3} \right)}^2} + {{\left( {m - 5} \right)}^2}} \ \Rightarrow {\left( {l + m} \right)^2} = 2\left[ {{{\left( {l - 3} \right)}^2} + {{\left( {m - 5} \right)}^2}} \right]\ \Rightarrow {\left( {x + y} \right)^2} = 2\left[ {{{\left( {x - 3} \right)}^2} + {{\left( {y - 5} \right)}^2}} \right]\end{array}$

eccentricity of the conic $25x^2-9y^2 = 225$, are

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

  1. $\dfrac25$

  2. $\dfrac45$

  3. $\dfrac13$

  4. $\dfrac15$

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Correct Option: B

The equation $(13x - 1)^{2} + (13y - 1)^{2} = k(5x - 12y + 1)^{2}$ will represent a parabola if

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

  1. $k = 2$

  2. $k = 81$

  3. $k = 169$

  4. $k = 1$

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Correct Option: D
Explanation:
We know that the general equation of the ellipse is 
${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}=e\dfrac{{\left(lh+mk+n\right)}^{2}}{{l}^{2}+{m}^{2}}$        ................$(1)$

${\left(13x-1\right)}^{2}+{\left(13y-1\right)}^{2}=k{\left(5x-12y+1\right)}^{2}$

$\Rightarrow\,{13}^{2}{\left(x-\dfrac{1}{13}\right)}^{2}+{13}^{2}{\left(y-\dfrac{1}{13}\right)}^{2}=k{\left(5x-12y+1\right)}^{2}$

$\Rightarrow\,169\left[{\left(x-\dfrac{1}{13}\right)}^{2}+{\left(y-\dfrac{1}{13}\right)}^{2}\right]=k{\left(5x-12y+1\right)}^{2}$

$\Rightarrow\,{\left(x-\dfrac{1}{13}\right)}^{2}+{\left(y-\dfrac{1}{13}\right)}^{2}=\dfrac{k}{169}{\left(5x-12y+1\right)}^{2}$       ...........$(2)$


A parabola has its eccentricity $e=1$

Comparing equations $(1)$ and $(2)$ we get

$\Rightarrow\,k=1$

$\therefore\,k=1$