Tag: mathematics and statistics

Questions Related to mathematics and statistics

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

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

  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 $3$

Show answer
Correct Option: A
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 $3$

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

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

  1. $(0,0)$

  2. $\left(\dfrac {1}{2},\dfrac {1}{4}\right)$

  3. $\left(-\dfrac {1}{4},\dfrac {1}{8}\right)$

  4. $\left(-\dfrac {1}{4},0\right)$

Show answer
Correct Option: D
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 $\left(\dfrac{-1}{4},0\right)$

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

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

  1. $3x+2y+14=0$

  2. $3x+2y-25=0$

  3. $2x-3y+10=0$

  4. none of these

Show answer
Correct Option: A
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

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

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

  1. $\displaystyle x=5 $

  2. $\displaystyle y=5 $

  3. $\displaystyle x=3 $

  4. $\displaystyle x=-3 $

Show answer
Correct Option: C
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.

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

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

  1. $(-3, -2)$

  2. $(-3,2)$

  3. $(3, -2)$

  4. None of these.

Show answer
Correct Option: C
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

 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

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

  1. $\displaystyle 12x+5y+20=0 $

  2. $\displaystyle 12x- 5y+17 =0 $

  3. $\displaystyle 12x+5y+17 =0 $

  4. $\displaystyle 12x+5y-27=0 $

Show answer
Correct Option: D
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 $\displaystyle 12x+5y-27=0$

The equation of directrix from the following is,

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

  1. $2x - y = 0$

  2. $x + 2y = 0$

  3. $x + y = 0$

  4. $x + 3y = 0$

Show answer
Correct Option: B
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.

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   

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

  1. 9x - 63y -2 = 0

  2. 59x -63y - 8 = 0

  3. 63x - 59y + 8 = 0

  4. 63x - 9y +2 = 0

Show answer
Correct Option: A

Asymptotes of the hyperbola $xy=4x+3y$ are 

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. x=3, y=4

  2. x=4, y=3

  3. x=2, y=6

  4. x=6, y=2

Show answer
Correct Option: A
Explanation:

correct answer is A.

Given,

$xy=4x+3y$
$\Rightarrow (x-y)=3(y-4)+12$
$\Rightarrow (x-4)(y-4)=12$
$\therefore$ joint equation of asymptotes is $(x-3)(y-4)=0$
hence,
$x-3=0$
$x=3$
and,
$y-4=0$

$y=4$

The angle between the asymptotes to the hyperbola $\dfrac { { x }^{ 2 } }{ 16 } -\dfrac { { y }^{ 2 } }{ 9 } =1$ is

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. $\pi -2\tan ^{ -1 }{ \left( \dfrac { 3 }{ 4 } \right) } $

  2. $\pi -2\tan ^{ -1 }{ \left( \dfrac { 4 }{ 3 } \right) } $

  3. $2\tan ^{ -1 }{ \left( \dfrac { 3 }{ 4 } \right) } $

  4. $2\tan ^{ -1 }{ \left( \dfrac { 4 }{ 3 } \right) } $

Show answer
Correct Option: C
Explanation:

Fact: Angle between asymptotes of hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ is $2\tan^{-1}\left(\dfrac{b}{a}\right)$


Hence angle between the asymptotes to the hyperbola $\dfrac{x^2}{16}-\dfrac{y^2}{9}=1$ is $2\tan^{-1}\left(\dfrac{3}{4}\right)$