Answer
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Hint:First we have to find the intersection point of chord and parabola . Put the value of $x$ or $y$ from the equation of chord to the equation of parabola and solve the quadratic equation by which we get two point of intersection and calculate the distance between the point
Complete step-by-step answer:
In this case firstly we have to find the point of intersection of chord and parabola ,
It is simple done by the solving the equation of parabola ${y^2} = 4x$ and chord $x + y = 1$
or $x = 1 - y$ , Putting the value of $x$ in equation of parabola ;
i.e. ${y^2} = 4\left( {1 - y} \right)$
by rearranging
${y^2} + 4y - 4$ = $0$
Now we have to solve this quadratic equation
\[a = 1\]
$b = 4$
$c = - 4$
therefore ,
$y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
by putting the values
$y = \dfrac{{ - 4 \pm \sqrt {{{\left( 4 \right)}^2} - 4 \times 1 \times \left( { - 4} \right)} }}{{2 \times 1}}$
After further solving ;
$y = \dfrac{{ - 4 \pm \sqrt {32} }}{2}$
$y = - 2 \pm 2\sqrt 2 $
It means that $y = - 2 + 2\sqrt 2 , - 2 - 2\sqrt 2 $
Now we have to put these $y$ values in equation of chord to get $x$ ;
Equation of chord is $x = 1 - y$ ;
therefore $x = 1 - ( - 2 + 2\sqrt 2 )$ or $x = 1 - ( - 2 - 2\sqrt 2 )$
we get $x = 3 - 2\sqrt 2 ,3 + 2\sqrt 2 $
So the point of intersection is $(3 - 2\sqrt 2 , - 2 + 2\sqrt 2 )$ and $(3 + 2\sqrt 2 , - 2 - 2\sqrt 2 )$
Now we have to calculate distance between them by using distance formula i.e. $\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
= $\sqrt {{{(3 + 2\sqrt 2 - 3 + 2\sqrt 2 )}^2} + {{( - 2 - 2\sqrt 2 + 2 - 2\sqrt 2 )}^2}} $
After solving we get
$ = \sqrt {4 \times 4 \times 2 + 4 \times 4 \times 2} $
$ = \sqrt {64} $
$ = 8 $
So,the length of the chord intercepted by the parabola ${y^2} = 4x$ on the straight line $x + y = 1$ is $8$
So, the correct answer is “Option C”.
Note:You can also simplify this question by putting the value of $y$ in the equation of parabola and get the $x$.If we get only one point of intersection then the chord is tangent of the parabola .
Complete step-by-step answer:
In this case firstly we have to find the point of intersection of chord and parabola ,
It is simple done by the solving the equation of parabola ${y^2} = 4x$ and chord $x + y = 1$
or $x = 1 - y$ , Putting the value of $x$ in equation of parabola ;
i.e. ${y^2} = 4\left( {1 - y} \right)$
by rearranging
${y^2} + 4y - 4$ = $0$
Now we have to solve this quadratic equation
\[a = 1\]
$b = 4$
$c = - 4$
therefore ,
$y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
by putting the values
$y = \dfrac{{ - 4 \pm \sqrt {{{\left( 4 \right)}^2} - 4 \times 1 \times \left( { - 4} \right)} }}{{2 \times 1}}$
After further solving ;
$y = \dfrac{{ - 4 \pm \sqrt {32} }}{2}$
$y = - 2 \pm 2\sqrt 2 $
It means that $y = - 2 + 2\sqrt 2 , - 2 - 2\sqrt 2 $
Now we have to put these $y$ values in equation of chord to get $x$ ;
Equation of chord is $x = 1 - y$ ;
therefore $x = 1 - ( - 2 + 2\sqrt 2 )$ or $x = 1 - ( - 2 - 2\sqrt 2 )$
we get $x = 3 - 2\sqrt 2 ,3 + 2\sqrt 2 $
So the point of intersection is $(3 - 2\sqrt 2 , - 2 + 2\sqrt 2 )$ and $(3 + 2\sqrt 2 , - 2 - 2\sqrt 2 )$
Now we have to calculate distance between them by using distance formula i.e. $\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
= $\sqrt {{{(3 + 2\sqrt 2 - 3 + 2\sqrt 2 )}^2} + {{( - 2 - 2\sqrt 2 + 2 - 2\sqrt 2 )}^2}} $
After solving we get
$ = \sqrt {4 \times 4 \times 2 + 4 \times 4 \times 2} $
$ = \sqrt {64} $
$ = 8 $
So,the length of the chord intercepted by the parabola ${y^2} = 4x$ on the straight line $x + y = 1$ is $8$
So, the correct answer is “Option C”.
Note:You can also simplify this question by putting the value of $y$ in the equation of parabola and get the $x$.If we get only one point of intersection then the chord is tangent of the parabola .
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