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

The ratio in which the line segment joining the points $(4, -6)$ and $(3, 1)$ is divided by the parabola $y^2 = 4x$ is

  1. $\displaystyle \frac{-20 \pm \sqrt{155}}{11}: 1$
  2. $\displaystyle \frac{-2 \pm 2\sqrt{155}}{11}: 2$
  3. $-20 \pm 2 \sqrt{155} : 11$
  4. $- 20 \pm \sqrt{155} : 11$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let P(h,k) be a point on the parabola which divides the line segment joining the points A(4,-6) and B(3,1) in the ratio $\lambda:1$.
So, coordinates of point P is $\displaystyle(\frac{3\lambda+4}{\lambda+1},\frac{\lambda-6}{\lambda+1}) $
But this point lies on the parabola
$\displaystyle(\frac{\lambda-6}{\lambda+1})^{2}=4(\frac{3\lambda+4}{\lambda+1})$
$\Rightarrow (\lambda-6)^{2}=4(3\lambda+4)(\lambda+1)$
$\Rightarrow 11\lambda^{2}+40\lambda-20=0$
$\Rightarrow \displaystyle \lambda=\frac{-20\pm 2\sqrt{155}}{11}$
So, the ratio will be ${-20\pm 2\sqrt{155}}:11$

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

Each member of the family of parabolas $y=ax^2+2x+3$ has a maximum or a minimum point depending upon the value of $a$. The equation of the locus of the maxima or minima for all possible values of $a$ is

  1. a straight line with slope $1$ and $y$ intercept $3$
  2. a straight line with slope $2$ and $y$ intercept $2$
  3. a straight line with slope $1$ and $x$ intercept $3$
  4. a straight line with slope $2$ and $y$ intercept $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Solution:- (A) a straight line with slope $1$ and $y$-intercept $3$
Consider the general case 
$y = a{x}^{2} + bx + c ..... \left( 1 \right)$. 
Given that the turning point (maximum or minimum) is on the axis of symmetry of the parabola, whose equation is $x = - \cfrac{b}{2a}$, 
$\therefore x$-coordinate of the turning point $= - \cfrac{b}{2a}$.
For $y$-coordinate of the turning point-
Substituting $x = -\cfrac{b}{2a}$ in ${eq}^{n} \left( 1 \right)$, we have

$y = a {\left( - \cfrac{b}{2a} \right)}^{2} + b \left( -\cfrac{b}{2a} \right) + c$
$y = \cfrac{{b}^{2}}{4a} - \cfrac{{b}^{2}}{a} + c$
$\Rightarrow y = -\cfrac{{b}^{2}}{4a} + c$
$\Rightarrow y = \cfrac{bx}{2} + c ..... \left( 2 \right) \; \left[ \text{independent of a} \right]$
Given equation of parabola-
$y = a{x}^{2} + 2x + 3$
Here
$b = 2 \; & \; c = 3$
Substituting these values in ${eq}^{n} \left( 2 \right)$, we get
$y = \cfrac{2x}{2} + 3$
$\Rightarrow y = x + 3$
Hence the equation of the locus of the maxima or minima for all possible values of a is a straight line with slope $1$ and $y$-intercept $3$

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

The focus of the parabola $y=2x^{2}+x$ is

  1. $(0,0)$
  2. $\left(\dfrac {1}{2},\dfrac {1}{4}\right)$
  3. $\left(-\dfrac {1}{4},\dfrac {1}{8}\right)$
  4. $\left(-\dfrac {1}{4},0\right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The given equation of parabola is 
$y=2{x}^{2}+x\Rightarrow {x}^{2}+\dfrac{x}{2}=\dfrac{y}{2}$
$\Rightarrow {x}^{2}+2\times \dfrac{x}{2}\times  \dfrac{1}{2}+\dfrac{1}{4}=\dfrac{y}{2}+\dfrac{1}{4}$
$\Rightarrow {\left(x+\dfrac{1}{2}\right)}^{2}=\dfrac{1}{2}\left(y+\dfrac{1}{8}\right)$
is of the form ${X}^{2}=\dfrac{1}{2}Y$ ......$(1)$    where $A=\dfrac{1}{8}$
Focus of $(1)$ is $\left(0,\dfrac{1}{8}\right)$ 
 where $X=0,Y=\dfrac{1}{8}$
$\Rightarrow x+\dfrac{1}{4}=0$ and $y+\dfrac{1}{8}=\dfrac{1}{8}$
$\Rightarrow x=\dfrac{-1}{4}$ and $y=0$
$\therefore$ focus of given parabola is $\left(\dfrac{-1}{4},0\right)$
Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

If the vertex and the focus of a parabola are $\left (-1,1 \right )$ and $\left (2,3 \right )$ respectively, then the equation of the directrix is

  1. $3x+2y+14=0$
  2. $3x+2y-25=0$
  3. $2x-3y+10=0$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

slope of axis $= \dfrac{3-1}{2+1} = \dfrac{2}{3}$
Slope of directrix $= \dfrac{-3}{2}$
Vertex is midpoint of foot of directrix and focus thus we get the coordinates oif foot of directrix as $(-4,-1)$
Equation of directrix will be $\dfrac{-3}{2}$  $= \dfrac{y+1}{x+4}$

Option A

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

The axis of the conic $\displaystyle x^{2}+4y-6x+17=0$ is

  1. $\displaystyle x=5 $
  2. $\displaystyle y=5 $
  3. $\displaystyle x=3 $
  4. $\displaystyle x=-3 $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x^2 + 4y -6x +17 =0$

$\Rightarrow x^2 -6x = -4y-17$
$\Rightarrow x^2 - 6x + 9 = -4y + 8$
$\Rightarrow (x-3)^2 = -4(y-2)$
Let $X = x-3$ and $Y = y-2$, then we get
$X^2 = -4Y$
Comparing the above equation with the standard equation of the parabola, we get that the axis of the parabola is given by $X=0$
Hence, axis of the given parabola will be $x-3 = 0\Rightarrow x =3$
Option $C$ is correct.

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 vertex of the parabola is

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

  4. None of these.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As the conic $\displaystyle x^{2}+4y-6x+k=0$ and $L=y+1=0$        .............( * )


$\displaystyle \Rightarrow x^{2}-6x= -4y-k$

$\displaystyle \Rightarrow \left ( x-3 \right )^{2}=-4y-k+9=-4\left ( y+\frac{k-9}{4} \right )$

$\displaystyle \Rightarrow \left ( x-3 \right )^{2}=-4y=4(-1)y$

$\displaystyle \therefore $ Equation of directrix is $y=1$

$\displaystyle \Rightarrow y+\frac{k-9}{4}=1$

$\displaystyle \Rightarrow y=1-\frac{k-9}{4}=\frac{13-k}{4}$          .....( **)

According to the problem (data given)

$\dfrac{13-k}{4}=-1\Rightarrow k =17$

So the equation of conic is

$\displaystyle \left ( x-3 \right )^{2}=-4\left ( y+2 \right ) \ \ \ \left ( \because k=17 \right )$

$\displaystyle X^{2}=-4(Y)=4(-1)Y$

$\displaystyle \Rightarrow $ whose vertex  is $ X=0$ & $ Y=0$

$\displaystyle \Rightarrow x-3=0 \ and \ y+2=0$

$\displaystyle \Rightarrow x=3 , y=-2$

$\displaystyle \therefore $ vertex $(3,-2)$

For $k=17$ from ( * *) requation of directrix is $y = -1$

Hence choice (c) is correct answer

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 equation of axis of the conic $()$  is

  1. $\displaystyle 12x+5y+20=0 $
  2. $\displaystyle 12x- 5y+17 =0 $
  3. $\displaystyle 12x+5y+17 =0 $
  4. $\displaystyle 12x+5y-27=0 $
Reveal answer Fill a bubble to check yourself
D 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 directrix' & the
constant ratio' is known as eccentricity denoted by $e$.
$\displaystyle \therefore e=PS/PM$
Now $\displaystyle 169\left { \left ( x-1 \right )^{2}+\left ( y-3 \right )^{2} \right }=\left ( 5x-12y+17 \right )^{2}$
$\displaystyle\Rightarrow \left { \left ( x-1 \right )^{2}+\left ( y-3 \right )^{2}
\right }=\frac{\left ( 5x-12y+17 \right )^{2}}{5^{2}+(-12)^{2}}=\left (
\frac{5x-12y+17}{13} \right )^{2}$
$\displaystyle \therefore \left
( x-1 \right )^{2}+\left ( y-3 \right )^{2}=e^{2}\left (
\frac{5x-12y+17}{13} \right )^{2}$ where $e=1$
Any line passing through focus $(1, 3)$ and perpendicular to $
\displaystyle 5x-12y+17=0$ is the axis of conic (*). Now any line
$\displaystyle \perp$er to $ \displaystyle 5x-12y+17=0$ is given by
$\displaystyle 12x+5y+\lambda =0$ but it passes through focus
$\displaystyle \therefore \lambda =-12\left ( 1 \right )-5\left ( 3 \right )=-27$
$\displaystyle \therefore $ equation of axis is $\displaystyle 12x+5y-27=0$

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

The equation of directrix from the following is,

  1. $2x - y = 0$
  2. $x + 2y = 0$
  3. $x + y = 0$
  4. $x + 3y = 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let $(a,b)$ be the focus and $y=m{x}$ be directrix 

$\implies \bigg(\dfrac{m{x}-y}{\sqrt{1+m^{2}}}\bigg)^{2}=(x-a)^{2}+(y-b)^{2}$

Differentiating on both sides

$\dfrac{(m{x}-y)(m-\dfrac{d{y}}{d{x}})}{1+m^{2}}=(x-a)+(y-b)\dfrac{d{y}}{d{x}}$

$x$ axis is tangent at $(1,0)$

$\dfrac{m(m)}{1+m^{2}}=1-a\implies a=\dfrac{1}{1+m^{2}}$

$y$ axis  is tangent at $(0,2)$

$\dfrac{-2}{1+m^{2}}=b-2\implies b=\dfrac{2{m}^{2}}{1+m^{2}}$

$(1,0)$ lies on parabola

$\dfrac{(m)^{2}}{1+m^{2}}=(1-a)^{2}+b^{2}$

Substituting $a$ and $b$ values 

$\implies (4{m^2}-1)(m^{2})=0\implies m=0,\pm \dfrac{1}{2}$

For $m=0$ we get $a=1,b=0$ which means that the directrix cuts the parabola which is not possible so $m=\pm \dfrac{1}{2}$

$\implies a=\dfrac{4}{5},b=\dfrac{2}{5}$

So the focus is $\bigg(\dfrac{4}{5},\dfrac{2}{5}\bigg)$

the directrix is $2{y}+x=0$

Hence option $B$ is the answer.
Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

The equation of pair of tangents to a parabola is given by $3x^2 +4y^2 +7xy -2x -y - 5 =0 $ and its focus is (1, 1), then the equation of directrix of the parabola is given by   

  1. 9x - 63y -2 = 0

  2. 59x -63y - 8 = 0

  3. 63x - 59y + 8 = 0

  4. 63x - 9y +2 = 0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation of the pair of tangents is given. The directrix of a parabola is the polar of the focus with respect to the parabola. Using the properties of the pair of tangents and the focus, the directrix is calculated as 9x - 63y - 2 = 0.