Tag: intersection of a line and a parabola

Questions Related to intersection of a line and a parabola

Let the equation of a circle and a parabola be $x^2+y^2-4x-6=0$ and $y^2=9x$ respectively. Then

intersection of a line and a parabola conic section maths

  1. $\left ( 1,-1 \right )$ is a point on the common chord of contact

  2. the equation of the common chord is $y+1=0$

  3. the length of the common chord is $6$

  4. none of these

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Correct Option: A,C
Explanation:

Given parabola and circle are $y^2=9x$ and $x^2+y^2-4x-6=0$ respectively.
Now solving these, $x^2+9x-4x-6=0\Rightarrow x^2+5x-6=0\Rightarrow (x-1)(x+6)=0\Rightarrow x = 1,-6$ but $x < 0$ is not a solution
Thus $x = 1, $ and corresponding $y = 3,-3$
Hence point of intersection of the parabola and circle are $(1,-3), (1,3)$
Hence equation of common chord is given by, $x=1$
and length of common chord $=3-(-3)=6$

The condition that the straight line $\displaystyle lx + my + n = 0$ touches the parabola $\displaystyle x^2 = 4ay$ is

intersection of a line and a parabola conic section maths

  1. $\displaystyle bn = am^2$

  2. $\displaystyle al^2 - mn = 0$

  3. $\displaystyle ln = am^2$

  4. $\displaystyle am = ln^2$

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Correct Option: B

The length of the chord of the parabola $y^2 = x$ which is bisected at the point $(2, 1)$ is

intersection of a line and a parabola conic section maths

  1. $2 \sqrt{3}$

  2. $4 \sqrt{3}$

  3. $3 \sqrt{2}$

  4. $2 \sqrt{5}$

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Correct Option: D
Explanation:

Chord through $(2, 1)$ is $ \cfrac{x- 2}{\cos  \theta} = \cfrac{y - 1}{\sin \theta} = r$    ... (i)
Solving equation (i) with parabola $y^2 = x$, we have
$(1 + r  \sin \theta)^2 = 2 + r  \cos \theta$
$\Rightarrow \sin^2 \theta r^2 + (2  \sin  \theta - \cos  \theta) r - 1 = 0$
This equation has two roots $r _1 = AC$ and $r _2 = - BC$
Then, sum of roots $r _1 + r _2 = 0$
$\Rightarrow 2\sin \theta - \cos  \theta = 0   \Rightarrow   \tan  \theta = \cfrac{1}{2}$
$AB = |r _1 - r _2| $
$= \sqrt{(r _1 + r _2)^2 - 4r _1 r _2}$
$= \sqrt{ 4 \cfrac{1}{\sin^2 \theta}} $

$= 2 \sqrt{5}$

If $2$ and $3$ are the length of the segments of any focal chord of a parabola $y^2 = 4ax$, then value of $2a$ is

intersection of a line and a parabola conic section maths

  1. $\dfrac{13}{5}$

  2. $\dfrac{12}{5}$

  3. $\dfrac{11}{5}$

  4. none of these

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Correct Option: B
Explanation:

If $2$ and $3$ are lengths of focal chord of parabola ${ y }^{ 2 }=4ax$ then 2a is :

Let ${ l } _{ 1 }=2\quad { l } _{ 2 }=3$
Semi latus rectum$=2a=\cfrac { 2{ l } _{ 1 }{ l } _{ 2 } }{ { l } _{ 1 }+{ l } _{ 2 } } $
$=\cfrac { 2(2)(3) }{ 2+3 } =\cfrac { 12 }{ 5 } $

If the line $y- \sqrt x +3 = 0$ cuts the parabola $y^2 = x + 2$ at $A$ and $B$, and if $P$ $(3,\ 0)$, then $PA.PB$ is equal to

intersection of a line and a parabola conic section maths

  1. $\dfrac{2(\sqrt 3+2)}{3}$

  2. $\dfrac{4\sqrt 3}{2}$

  3. $\dfrac{4(2-\sqrt 3)}{3}$

  4. $\dfrac{4(\sqrt3+2)}{3}$

Show answer
Correct Option: D
Explanation:

If line  $y-\sqrt { 3 } x+3=0$ cuts parabola ${ y }^{ 2 }=x+2$ at $A$ and $B$ 

If $P(3,0)$ the $PA.PB$ is : 
$y-\sqrt { 3 } x+3=0\ y=\sqrt { 3 } x-3\ \cfrac { y-0 }{ \cfrac { \sqrt { 3 }  }{ 2 }  } =\cfrac { x-\sqrt { 3 }  }{ \cfrac { 1 }{ 2 }  } =r\ x=\cfrac { r }{ 2 } +\sqrt { 3 } \ y=\cfrac { \sqrt { 3 } r }{ 2 } $
Put in ${ y }^{ 2 }=x+2$
$\cfrac { 3 }{ 4 } { r }^{ 2 }=\cfrac { r }{ 2 } +\sqrt { 3 } +2\ { 3r }^{ 2 }-2r-4(\sqrt { 3 } +2)=0\ PA.PB=\left| { r } _{ 1 }{ r } _{ 2 } \right| =\left| \cfrac { c }{ a }  \right| =\cfrac { 4(\sqrt { 3 } +2) }{ 3 } $