Tag: parabola

Questions Related 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

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  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. $- 20 \pm \sqrt{155} : 11$

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Correct Option: C
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 ${-20\pm 2\sqrt{155}}:11$

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

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  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 $3$

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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

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  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

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  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

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  1. $\displaystyle x=5 $

  2. $\displaystyle y=5 $

  3. $\displaystyle x=3 $

  4. $\displaystyle x=-3 $

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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

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  1. $(-3, -2)$

  2. $(-3,2)$

  3. $(3, -2)$

  4. None of these.

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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

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  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,

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  1. $2x - y = 0$

  2. $x + 2y = 0$

  3. $x + y = 0$

  4. $x + 3y = 0$

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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   

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  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