Tag: intersection of a line and a parabola
Questions Related to intersection of a line and a parabola
The length of the chord of the parabola $y^2 = 4x$ which passes through the vertex and makes $30^o$ angle with x-axis is
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$\dfrac{\sqrt{3}}{2}$
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$\dfrac{3}{2}$
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$8\sqrt{3}$
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$\sqrt{3}$
If a$\ne $b then the length of common chord of the circles ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {c^2}$ and ${\left( {x - {b^{}}} \right)^2} + {\left( {y - a} \right)^2} = c^2$ is
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$\sqrt {{c^2} - {{\left( {a - b} \right)}^2}} $
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$\sqrt {4{c^2} - 2{{\left( {a - b} \right)}^2}} $
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$\sqrt {3{c^2} - {{\left( {a - b} \right)}^2}} $
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$\sqrt {2{c^2} - {{\left( {a - b} \right)}^2}} $
We have,
${{\left(
x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{c}^{2}}$
${{S} _{1}}\equiv
{{x}^{2}}+{{a}^{2}}-2ax+{{y}^{2}}+{{b}^{2}}-2by-{{c}^{2}}=0$
…….. (1)
${{\left( x-b \right)}^{2}}+{{\left( y-a \right)}^{2}}={{c}^{2}}$
${{S} _{2}}\equiv {{x}^{2}}+{{b}^{2}}-2bx+{{y}^{2}}+{{a}^{2}}-2ay-{{c}^{2}}=0$ ……… (2)
Since, $a\ne b$
Centre of the circle ${{S} _{1}}=\left( a,b \right)$ and radius ${{r} _{1}}=c$.
We know that the equation of common chord is ${{S} _{1}}-{{S} _{2}}=0$
So, the equation is
$\left( b-a \right)x+\left( a-b \right)y=0$ …….. (3)
We know that the length of common chord is
$=2\sqrt{{{r} _{1}}^{2}-{{d} _{1}}^{2}}$ ………. (4)
Where ${{r} _{1}}=$ radius and ${{d} _{1}}$ is the length of perpendicular drawn from the centre to the chord.
So,
${{d} _{1}}=\left| \dfrac{\left( b-a \right)a+\left( a-b \right)b}{\sqrt{{{\left( b-a \right)}^{2}}+{{\left( a-b \right)}^{2}}}} \right|$
$ {{d} _{1}}=\left| \dfrac{ab-{{a}^{2}}+ab-{{b}^{2}}}{\sqrt{{{\left( a-b \right)}^{2}}+{{\left( a-b \right)}^{2}}}} \right| $
$ {{d} _{1}}=\left| \dfrac{2ab-{{a}^{2}}-{{b}^{2}}}{\sqrt{2{{\left( a-b \right)}^{2}}}} \right| $
$ {{d} _{1}}=\left| \dfrac{-{{\left( a-b \right)}^{2}}}{\sqrt{2{{\left( a-b \right)}^{2}}}} \right| $
$ {{d} _{1}}=\left| \dfrac{-\left( a-b \right)}{\sqrt{2}} \right| $
$ {{d} _{1}}=\dfrac{\left( a-b \right)}{\sqrt{2}} $
From equation (4),
The length of common chord $ =2\sqrt{{{c}^{2}}-{{\left( \dfrac{a-b}{\sqrt{2}} \right)}^{2}}} $
$ =2\sqrt{{{c}^{2}}-{{\dfrac{\left( a-b \right)}{2}}^{2}}} $
$ =2\sqrt{{{\dfrac{2{{c}^{2}}-\left( a-b \right)}{2}}^{2}}} $
$ =\sqrt{{{\dfrac{8{{c}^{2}}-4\left( a-b \right)}{2}}^{2}}} $
$ =\sqrt{4{{c}^{2}}-2{{\left( a-b \right)}^{2}}} $
Hence, this is the answer.
Length of chord of parabola ${y}^{2}=4ax$ whose equation is $y-\sqrt {2}x+4\sqrt {2}a=0$
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$2\sqrt {11}a$
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$4\sqrt {2}a$
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$8\sqrt {2}a$
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$6\sqrt {3}a$
If a$\ne $b then the length of common chord of the circles ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {c^2}$ and ${\left( {x - {b^{}}} \right)^2} + {\left( {y - a} \right)^2} = c^2$ is
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$\sqrt {{c^2} - {{\left( {a - b} \right)}^2}} $
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$\sqrt {3{c^2} - {{\left( {a - b} \right)}^2}} $
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$\sqrt {4{c^2} - 2{{\left( {a - b} \right)}^2}} $
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$\sqrt {2{c^2} - {{\left( {a - b} \right)}^2}} $
We have,
${{\left(
x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{c}^{2}}$
${{S} _{1}}\equiv
{{x}^{2}}+{{a}^{2}}-2ax+{{y}^{2}}+{{b}^{2}}-2by-{{c}^{2}}=0$
…….. (1)
${{\left( x-b \right)}^{2}}+{{\left( y-a \right)}^{2}}={{c}^{2}}$
${{S} _{2}}\equiv {{x}^{2}}+{{b}^{2}}-2bx+{{y}^{2}}+{{a}^{2}}-2ay-{{c}^{2}}=0$ ……… (2)
Since, $a\ne b$
Centre of the circle ${{S} _{1}}=\left( a,b \right)$ and radius ${{r} _{1}}=c$.
We know that the equation of common chord is ${{S} _{1}}-{{S} _{2}}=0$
So, the equation is
$\left( b-a \right)x+\left( a-b \right)y=0$ …….. (3)
We know that the length of common chord is
$=2\sqrt{{{r} _{1}}^{2}-{{d} _{1}}^{2}}$ ………. (4)
Where ${{r} _{1}}=$ radius and ${{d} _{1}}$ is the length of perpendicular drawn from the centre to the chord.
So,
${{d} _{1}}=\left| \dfrac{\left( b-a \right)a+\left( a-b \right)b}{\sqrt{{{\left( b-a \right)}^{2}}+{{\left( a-b \right)}^{2}}}} \right|$
$ {{d} _{1}}=\left| \dfrac{ab-{{a}^{2}}+ab-{{b}^{2}}}{\sqrt{{{\left( a-b \right)}^{2}}+{{\left( a-b \right)}^{2}}}} \right| $
$ {{d} _{1}}=\left| \dfrac{2ab-{{a}^{2}}-{{b}^{2}}}{\sqrt{2{{\left( a-b \right)}^{2}}}} \right| $
$ {{d} _{1}}=\left| \dfrac{-{{\left( a-b \right)}^{2}}}{\sqrt{2{{\left( a-b \right)}^{2}}}} \right| $
$ {{d} _{1}}=\left| \dfrac{-\left( a-b \right)}{\sqrt{2}} \right| $
$ {{d} _{1}}=\dfrac{\left( a-b \right)}{\sqrt{2}} $
From equation (4),
The length of common chord $ =2\sqrt{{{c}^{2}}-{{\left( \dfrac{a-b}{\sqrt{2}} \right)}^{2}}} $
$ =2\sqrt{{{c}^{2}}-{{\dfrac{\left( a-b \right)}{2}}^{2}}} $
$ =2\sqrt{{{\dfrac{2{{c}^{2}}-\left( a-b \right)}{2}}^{2}}} $
$ =\sqrt{{{\dfrac{8{{c}^{2}}-4\left( a-b \right)}{2}}^{2}}} $
$ =\sqrt{4{{c}^{2}}-2{{\left( a-b \right)}^{2}}} $
Hence, this is the answer.
The length of the chord $y = x - 2$ intercepted by the parabola ${ y }^{ 2 }=4(x-1)$ is
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$4$
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$\dfrac { 16 }{ 3 } $
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$\dfrac { 3 }{ 16 } $
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$\dfrac { 1 }{ 4 } $
The length of normal chord to the parabola $y^{2} = 4x$ which subtends a right angle at the vertex is
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$6\sqrt {3}$
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$6\sqrt {2}$
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$7\sqrt {2}$
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$7\sqrt {3}$
A chord is a line segment that passes through any two points on the parabola. A normal chord is a chord that is perpendicular to a tangent of the parabola at the point of intersection of the chord with the parabola.
The number of focal chord(s) of length $\dfrac{4}{7}$ in the parabola $7y^2 = 8x$ is
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$1$
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$0$
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infinite
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none of these
Parabola: ${ y }^{ 2 }=4(\cfrac { 2 }{ 7 } )x\quad ...........(1)\quad (a=\cfrac { 2 }{ 7 } )$
Find the length of the chord of the parabola $y^2\, =\, 8x$, whose equation is $x + y = 1$.
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$8 \sqrt{3}$
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$4 \sqrt {3}$
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$2 \sqrt {3}$
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$\sqrt {3}$
Substitute $x=1-y$ in $y^2=8x$
The length of the chord of the parabola $x^2 = 4y $ passing through the vertex and having slope $cot \alpha $ is
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$4 \cos \alpha . cosec^2\alpha$
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$ 4a \tan \alpha \sec \alpha $
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$4 \sin \alpha . \sec^2 \alpha $
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none of these
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
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$\sqrt{13}a$
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$4a$
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$5a$
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$2\sqrt{3}a$
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 $=\cfrac{4}{m^2}\sqrt{a(1+m^2)(a-cm)}=\sqrt{13}a$