Tag: parabola

Quizzes (1) Questions (49)

Questions Related to parabola

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

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

  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 $\left(\dfrac{3}{2}, 1\right)$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
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$=\left(\dfrac{1}{2}+1,1\right)$

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

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

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

  1. $y=-\dfrac {9}{4}, 8$
  2. $x=\dfrac {-9}{4}, 9$
  3. $y=\dfrac {9}{4}, 9$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.
Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

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

  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 $\displaystyle \left ( -4,0 \right )$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\displaystyle \left ( 4,0 \right ).$

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

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

  1. $12$
  2. $6$
  3. $3$
  4. $9$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $x+a=6$

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

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

  1. $y^2 = 4x$
  2. $y^2 = 5x$
  3. $y^2 = 6x$
  4. $y^2 = 8x$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the definition of a parabola (distance from focus equals distance to directrix), for focus (3/2, 0) and directrix x = -3/2, the equation is sqrt((x-3/2)^2 + y^2) = |x + 3/2|. Squaring both sides gives (x-3/2)^2 + y^2 = (x+3/2)^2, which simplifies to y^2 = 6x.

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

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

  1. $y^2=4ax$
  2. $y^2+4ax=0$
  3. $x^2+4ay=0$
  4. $x^2-4ay=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
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)


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

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

  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. $\begin{array}{}\\ \Rightarrow \left| \right| = {\left( {x + y} \right)^2} = 2\left[ {{{\left( {x - 6} \right)}^2} + {{\left( {y - 6} \right)}^2}} \right]\end{array}$<span class="mrow"><span class="mtable">
  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. $\begin{array}{}\\ \Rightarrow \left| \right| = {\left( {x + y} \right)^2} = 2\left[ {{{\left( {x - 7} \right)}^2} + {{\left( {y - 7} \right)}^2}} \right]\end{array}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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: $L:y+x=0$
$\begin{array}{l}QO = QS\ \Rightarrow \left| {\frac{{l + m}}{{\sqrt 2 }}} \right| = \sqrt {{{\left( {l - 3} \right)}^2} + {{\left( {m - 5} \right)}^2}} \ \Rightarrow {\left( {l + m} \right)^2} = 2\left[ {{{\left( {l - 3} \right)}^2} + {{\left( {m - 5} \right)}^2}} \right]\ \Rightarrow {\left( {x + y} \right)^2} = 2\left[ {{{\left( {x - 3} \right)}^2} + {{\left( {y - 5} \right)}^2}} \right]\end{array}$

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

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

  1. $k = 2$
  2. $k = 81$
  3. $k = 169$
  4. $k = 1$
Reveal answer Fill a bubble to check yourself
D Correct answer
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

$\Rightarrow\,k=1$

$\therefore\,k=1$