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Multiple choice intersection of a line and a parabola conic section maths

The number of focal chord(s) of length $\dfrac{4}{7}$ in the parabola $7y^2 = 8x$ is

  1. $1$
  2. $0$

  3. infinite

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

Parabola: ${ y }^{ 2 }=4(\cfrac { 2 }{ 7 } )x\quad ...........(1)\quad (a=\cfrac { 2 }{ 7 } )$

Let there be a point $P(a{ t }^2,2at)$ on parabola lying on focal chord so it will intersect parabola at $Q(\cfrac { a }{ t }^2,\cfrac { -2a }{ t })$(as ${ t } _{ 1 }{ t } _{ 2 }=-1)$ 
So, length of $PQ:\sqrt { (\cfrac { a }{ { t }^{ 2 } } -{ { at }^{ 2 }) }^{ 2 }+4(\cfrac { a }{ t } +{ at) }^{ 2 } } $
Given $PQ=\cfrac{ 4 }{ 7 }=\cfrac{ 2 }{ 7 }\sqrt { { (\cfrac { 1 }{ { t }^{ 2 } } -{ t }^{ 2 }) }^{ 2 }+4({ \cfrac { 1 }{ t } +t) }^{ 2 } } $
$=>4=(\cfrac { 1 }{ { t }^{ 2 } } -{ { t }^{ 2 }) }^{ 2 }+4(\cfrac { 1 }{ t } +{ t })^{ 2 }$
$=>4=[(\cfrac { 1 }{ t } +t)(\cfrac { 1 }{ t } -{ t)] }^{ 2 }+4(\cfrac { 1 }{ t } +{ t })^{ 2 }$
$=>4={ (\cfrac { 1 }{ t } +t) }^{ 2 }[({ \cfrac { 1 }{ t } -2) }^{ 2 }+4]$
$=>4=({ \cfrac { 1 }{ t } +t) }^{ 2 }(\cfrac { 1 }{ { t }^{ 2 } } +{ t }^{ 2 }+4-2)$
$=>4=({ \cfrac { 1 }{ t } +t) }^{ 2 }({ \cfrac { 1 }{ t } +t) }^{ 2 }$
$=>4=(\cfrac { 1 }{ t } +{ t) }^{ 4 }$
$=>0=(\cfrac { 1 }{ t } +{ t })^{ 2 }(\cfrac { 1 }{ t } +{ t })^{ 2 }-({ 2 })^{ 2 }$
$=>[(\cfrac { 1 }{ t } +{ t })^{ 2 }+2][(\cfrac { 1 }{ t } +{ t })^{ 2 }-2]=0$
We know, $[(\cfrac { 1 }{ t } +{ t })^{ 2 }+2]$ can never be zero so $[(\cfrac { 1 }{ t } +{ t })^{ 2 }-2]=0$
$=>(\cfrac{1}{t}+t-\sqrt2)(\cfrac{1}{t}+t+\sqrt2)=0$
$=>({t}^2-\sqrt2t+1)(t^2+\sqrt2t+1)=0$
$=> $either$ I:(t^2-\sqrt2t+1)=0\quad $or$\quad II:(t^2+\sqrt2t+1)=0$
$D _1=(-\sqrt2)^2-4(1)=-2 <0, $ No solution.
$D _2=(\sqrt2)^2-4(1)=-2 <0,$ No solution.
No such focal chord is possible.

Multiple choice intersection of a line and a parabola conic section maths

Find the length of the chord of the parabola $y^2\, =\, 8x$, whose equation is $x + y = 1$.

  1. $8 \sqrt{3}$
  2. $4 \sqrt {3}$
  3. $2 \sqrt {3}$
  4. $\sqrt {3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Substitute $x=1-y$ in $y^2=8x$

Then $y^2=8(1-y)$
$y=\dfrac{8\pm\sqrt{96}}{2}=4\pm 2\sqrt6$
Therefore, $x=3 \mp 2\sqrt6$
Using distance formula length of chord $=8\sqrt3$

Multiple choice intersection of a line and a parabola conic section maths

The length of the chord of the parabola $x^2 = 4y $ passing through the vertex and having slope $cot \alpha $ is 

  1. $4 \cos \alpha . cosec^2\alpha$
  2. $ 4a \tan \alpha \sec \alpha $
  3. $4 \sin \alpha . \sec^2 \alpha $

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Equation of parabola $x^2=4ay$
Vertex of parabola $(0)=(0, 0)$
Equation of chord passing through vertex $y-y _1=m(x-x _1)$

Here, $x _1=0$ and $y _1=0$
$y-0=m(x-0)$
$\therefore y=mx$

Now, slope$=m=\tan\alpha$
Let the chord intersect the parabola at $P(h, k)$
Since, point P lies on chord
$k=(\tan\alpha)h$ ………..$(1)$

Point P lies on parabola
$h^2=4ak$ ………$(2)$

Substituting equation $(1)$ in $(2)$
$\therefore h^2=4a(\tan\alpha \times h)$
$\therefore h^2(h-4a\tan\alpha)=0$
$\therefore h=0$
$h=4a\tan\alpha$

Substituting $h=4a\tan\alpha$ in equation $(2)$, we get
$\therefore (4a)^2\tan^2\alpha=4ak$
$\therefore k=4a\tan^2\alpha$
$\therefore p=(h, k)=(4a\tan\alpha, 4a\tan^2\alpha)$

Distance $OP=\sqrt{(4a\tan\alpha -0)^2+(4a\tan^2\alpha -0)}$
$=\sqrt{16a^2\tan^2\alpha +16a^2\tan^4\alpha}$
$=4a\tan\alpha\sqrt{1+\tan^2\alpha}$
$=4a\tan\alpha\sqrt{\sec^2\alpha}$
$=4a\tan\alpha \sec\alpha$.
Multiple choice intersection of a line and a parabola conic section maths

Let $AB$ be a chord of the parabola $y^{2}=4ax$.If the pole of $AB$ with respect to the parabola be $\left ( 2a,3a \right )$ then the length of $AB$ is 

  1. $\sqrt{13}a$
  2. $4a$
  3. $5a$
  4. $2\sqrt{3}a$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given pole of AB w.r.t the parabola $y^2=4ax$ is $(2a,3a)$
Thus the equation of chord $AB$ is given by $T=0$
$\Rightarrow y(2a)-2a(x+3a)=0$
$\Rightarrow y =\dfrac{2}{3}x+\dfrac{4}{3}a$
Clearly here $m = \dfrac{2}{3}, c = \dfrac{4}{3}a$
Therefore length of chord AB is $=\cfrac{4}{m^2}\sqrt{a(1+m^2)(a-cm)}=\sqrt{13}a$

Multiple choice intersection of a line and a parabola conic section maths

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

  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
Reveal answer Fill a bubble to check yourself
A,C Correct answer
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 $=3-(-3)=6$

Multiple choice intersection of a line and a parabola conic section maths

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

  1. $\displaystyle bn = am^2$
  2. $\displaystyle al^2 - mn = 0$
  3. $\displaystyle ln = am^2$
  4. $\displaystyle am = ln^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a line y = mx + c to touch x^2 = 4ay, the condition is c = -am^2. Rewriting lx + my + n = 0 as y = (-l/m)x - (n/m), we set c = -n/m and m_slope = -l/m. Thus, -n/m = -a(-l/m)^2 = -a(l^2/m^2). Multiplying by -m^2 gives nm = al^2, or al^2 - nm = 0.

Multiple choice intersection of a line and a parabola conic section maths

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

  1. $2 \sqrt{3}$
  2. $4 \sqrt{3}$
  3. $3 \sqrt{2}$
  4. $2 \sqrt{5}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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}} $

$= 2 \sqrt{5}$

Multiple choice intersection of a line and a parabola conic section maths

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

  1. $\dfrac{13}{5}$
  2. $\dfrac{12}{5}$
  3. $\dfrac{11}{5}$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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 } $

Multiple choice intersection of a line and a parabola conic section maths

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

  1. $\dfrac{2(\sqrt 3+2)}{3}$
  2. $\dfrac{4\sqrt 3}{2}$
  3. $\dfrac{4(2-\sqrt 3)}{3}$
  4. $\dfrac{4(\sqrt3+2)}{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 } $

Multiple choice conjugate hyperbola hyperbola conic section maths

The
eccentricity of the hyperbola whose asymptotes are $3x + 4y = 2{\text{ and }}4x - 3y + 5 = 0$  

  1. 1

  2. 2
  3. $\sqrt 2 $
  4. $\sqrt 3 $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The asymptotes of a hyperbola differ from the hyperbola equation only by a constant term. Using the family of lines representing the asymptotes, we can find the angle between them and subsequently compute the eccentricity as sqrt(2) for a rectangular hyperbola where asymptotes are perpendicular.