If the parabola is translated from its initial position to a new position by moving its vertex along the line...
$begingroup$
The parabola $y=4-x^2$ has vertex $P.$It intersects $x-$axis at $A$ and $B.$ If the parabola is translated from its initial position to a new position by moving its vertex along the line $y=x+4,$ so that it intersects $x-$axis at $B$ and $C$,then find the abscissa of $C$
$x^2=-(y-4)$
$A(2,0),B(-2,0)$
Let the new vertex is $(x_1,x_1+4)$,so new parabola is $(x-x_1)^2=-(y-x_1-4).......(1)$
As new parabola intersects the $x$ axis at $(-2,0)$ so putting $x=-2,y=0$ i got
$x^2+4x_1+4-x_1-4=0$
$x_1=0,-3$
New parabola is $(x+3)^2=-(y-1)$ But this new parabola is cutting the $x$ axis at $(-2,0),(-4,0)$ but the answer given for abscissa of $C$ is $8.$
quadratics conic-sections
asked Dec 7 '18 at 6:26
user984325user984325
246112
$endgroup$
add a comment |
$begingroup$
The parabola $y=4-x^2$ has vertex $P.$It intersects $x-$axis at $A$ and $B.$ If the parabola is translated from its initial position to a new position by moving its vertex along the line $y=x+4,$ so that it intersects $x-$axis at $B$ and $C$,then find the abscissa of $C$
$x^2=-(y-4)$
$A(2,0),B(-2,0)$
Let the new vertex is $(x_1,x_1+4)$,so new parabola is $(x-x_1)^2=-(y-x_1-4).......(1)$
As new parabola intersects the $x$ axis at $(-2,0)$ so putting $x=-2,y=0$ i got
$x^2+4x_1+4-x_1-4=0$
$x_1=0,-3$
New parabola is $(x+3)^2=-(y-1)$ But this new parabola is cutting the $x$ axis at $(-2,0),(-4,0)$ but the answer given for abscissa of $C$ is $8.$
quadratics conic-sections
asked Dec 7 '18 at 6:26
user984325user984325
246112
$endgroup$
add a comment |
$begingroup$
The parabola $y=4-x^2$ has vertex $P.$It intersects $x-$axis at $A$ and $B.$ If the parabola is translated from its initial position to a new position by moving its vertex along the line $y=x+4,$ so that it intersects $x-$axis at $B$ and $C$,then find the abscissa of $C$
$x^2=-(y-4)$
$A(2,0),B(-2,0)$
Let the new vertex is $(x_1,x_1+4)$,so new parabola is $(x-x_1)^2=-(y-x_1-4).......(1)$
As new parabola intersects the $x$ axis at $(-2,0)$ so putting $x=-2,y=0$ i got
$x^2+4x_1+4-x_1-4=0$
$x_1=0,-3$
New parabola is $(x+3)^2=-(y-1)$ But this new parabola is cutting the $x$ axis at $(-2,0),(-4,0)$ but the answer given for abscissa of $C$ is $8.$
quadratics conic-sections
asked Dec 7 '18 at 6:26
user984325user984325
246112
$endgroup$
The parabola $y=4-x^2$ has vertex $P.$It intersects $x-$axis at $A$ and $B.$ If the parabola is translated from its initial position to a new position by moving its vertex along the line $y=x+4,$ so that it intersects $x-$axis at $B$ and $C$,then find the abscissa of $C$
$x^2=-(y-4)$
$A(2,0),B(-2,0)$
Let the new vertex is $(x_1,x_1+4)$,so new parabola is $(x-x_1)^2=-(y-x_1-4).......(1)$
As new parabola intersects the $x$ axis at $(-2,0)$ so putting $x=-2,y=0$ i got
$x^2+4x_1+4-x_1-4=0$
$x_1=0,-3$
New parabola is $(x+3)^2=-(y-1)$ But this new parabola is cutting the $x$ axis at $(-2,0),(-4,0)$ but the answer given for abscissa of $C$ is $8.$
quadratics conic-sections
quadratics conic-sections
asked Dec 7 '18 at 6:26
user984325user984325
246112
asked Dec 7 '18 at 6:26
user984325user984325
246112
asked Dec 7 '18 at 6:26
user984325user984325
246112
asked Dec 7 '18 at 6:26
user984325user984325
246112
asked Dec 7 '18 at 6:26
user984325user984325
246112
246112
add a comment |
add a comment |
1 Answer
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$begingroup$
Note that the question just says that the parabola intersects the $x$ axis at points $A,B$. You considered the possibility $A(2,0),B(-2,0)$, but $A(-2,0),B(2,0)$ is another possibility. Both your and your book's answers are correct, but incomplete.
If we take $A(-2,0),B(2,0)$, the new parabola passes through $B(2,0)$ and has the equation $(x-5)^2=9-y$, which indeed cuts the $x$ axis at $(8,0)$.
answered Dec 7 '18 at 7:08
Shubham JohriShubham Johri
5,057717
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1 Answer
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1 Answer
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$begingroup$
Note that the question just says that the parabola intersects the $x$ axis at points $A,B$. You considered the possibility $A(2,0),B(-2,0)$, but $A(-2,0),B(2,0)$ is another possibility. Both your and your book's answers are correct, but incomplete.
If we take $A(-2,0),B(2,0)$, the new parabola passes through $B(2,0)$ and has the equation $(x-5)^2=9-y$, which indeed cuts the $x$ axis at $(8,0)$.
answered Dec 7 '18 at 7:08
Shubham JohriShubham Johri
5,057717
$endgroup$
add a comment |
$begingroup$
Note that the question just says that the parabola intersects the $x$ axis at points $A,B$. You considered the possibility $A(2,0),B(-2,0)$, but $A(-2,0),B(2,0)$ is another possibility. Both your and your book's answers are correct, but incomplete.
If we take $A(-2,0),B(2,0)$, the new parabola passes through $B(2,0)$ and has the equation $(x-5)^2=9-y$, which indeed cuts the $x$ axis at $(8,0)$.
answered Dec 7 '18 at 7:08
Shubham JohriShubham Johri
5,057717
$endgroup$
add a comment |
$begingroup$
Note that the question just says that the parabola intersects the $x$ axis at points $A,B$. You considered the possibility $A(2,0),B(-2,0)$, but $A(-2,0),B(2,0)$ is another possibility. Both your and your book's answers are correct, but incomplete.
If we take $A(-2,0),B(2,0)$, the new parabola passes through $B(2,0)$ and has the equation $(x-5)^2=9-y$, which indeed cuts the $x$ axis at $(8,0)$.
answered Dec 7 '18 at 7:08
Shubham JohriShubham Johri
5,057717
$endgroup$
Note that the question just says that the parabola intersects the $x$ axis at points $A,B$. You considered the possibility $A(2,0),B(-2,0)$, but $A(-2,0),B(2,0)$ is another possibility. Both your and your book's answers are correct, but incomplete.
If we take $A(-2,0),B(2,0)$, the new parabola passes through $B(2,0)$ and has the equation $(x-5)^2=9-y$, which indeed cuts the $x$ axis at $(8,0)$.
answered Dec 7 '18 at 7:08
Shubham JohriShubham Johri
5,057717
answered Dec 7 '18 at 7:08
Shubham JohriShubham Johri
5,057717
answered Dec 7 '18 at 7:08
Shubham JohriShubham Johri
5,057717
answered Dec 7 '18 at 7:08
Shubham JohriShubham Johri
5,057717
5,057717
add a comment |
add a comment |
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