
How do you find the vertex, focus and directrix of parabola ${{y}^{2}}-4y-4x=0$ ?
Answer
466.5k+ views
Hint: If the equation parabola is ${{\left( y-b \right)}^{2}}=4a\left( x-c \right)$ then the coordinate of vertex of parabola (c, b) the focus will be at a distance a units from the vertex along axis of parabola so the coordinate of focus is (c+a, b) since the axis of this parabola is parallel to x axis. The directrix is parallel to the tangent at vertex at a distance of units opposite to focus, so the equation of the directrix is $x=c-a$ . We will convert the equation given in the question to ${{\left( y-b \right)}^{2}}=4a\left( x-c \right)$ and then solve it.
Complete step by step answer:
The equation of given parabola is ${{y}^{2}}-4y-4x=0$
First we will convert the equation to ${{\left( y-b \right)}^{2}}=4a\left( x-c \right)$
${{y}^{2}}-4y-4x=0$
Adding 4 in LHS and RHS
$\Rightarrow {{y}^{2}}-4y+4-4x=4$
$\Rightarrow {{y}^{2}}-4y+4=4x+4$
$\Rightarrow {{\left( y-2 \right)}^{2}}=4\times 1\times \left( x+1 \right)$
If we compare the equation ${{\left( y-2 \right)}^{2}}=4\times 1\times \left( x+1 \right)$ with ${{\left( y-b \right)}^{2}}=4a\left( x-c \right)$ then the
$a=1$ , $b=2$ and $c=-1$
We know that if the equation parabola is ${{\left( y-b \right)}^{2}}=4a\left( x-c \right)$ the coordinate of vertex is (c,b)
The coordinate of the vertex of parabola ${{y}^{2}}-4y-4x=0$ is $\left( -1,2 \right)$
The focus is at distance of a units from vertex along the axis focus of parabola ${{\left( y-b \right)}^{2}}=4a\left( x-c \right)$ is $\left( c+a,b \right)$
So the coordinate of the focus of parabola ${{y}^{2}}-4y-4x=0$ is
$\left( 1+\left( -1 \right),2 \right)=\left( 0,2 \right)$
The directrix is parallel to tangent at vertex and at a distance of a units along the axis the equation of directrix of ${{\left( y-b \right)}^{2}}=4a\left( x-c \right)$ is $x=c-a$
So the equation of directrix of parabola ${{y}^{2}}-4y-4x=0$ is $x=-2$
If we draw the graph we can visualize it more
Complete step by step answer:
The equation of given parabola is ${{y}^{2}}-4y-4x=0$
First we will convert the equation to ${{\left( y-b \right)}^{2}}=4a\left( x-c \right)$
${{y}^{2}}-4y-4x=0$
Adding 4 in LHS and RHS
$\Rightarrow {{y}^{2}}-4y+4-4x=4$
$\Rightarrow {{y}^{2}}-4y+4=4x+4$
$\Rightarrow {{\left( y-2 \right)}^{2}}=4\times 1\times \left( x+1 \right)$
If we compare the equation ${{\left( y-2 \right)}^{2}}=4\times 1\times \left( x+1 \right)$ with ${{\left( y-b \right)}^{2}}=4a\left( x-c \right)$ then the
$a=1$ , $b=2$ and $c=-1$
We know that if the equation parabola is ${{\left( y-b \right)}^{2}}=4a\left( x-c \right)$ the coordinate of vertex is (c,b)
The coordinate of the vertex of parabola ${{y}^{2}}-4y-4x=0$ is $\left( -1,2 \right)$
The focus is at distance of a units from vertex along the axis focus of parabola ${{\left( y-b \right)}^{2}}=4a\left( x-c \right)$ is $\left( c+a,b \right)$
So the coordinate of the focus of parabola ${{y}^{2}}-4y-4x=0$ is
$\left( 1+\left( -1 \right),2 \right)=\left( 0,2 \right)$
The directrix is parallel to tangent at vertex and at a distance of a units along the axis the equation of directrix of ${{\left( y-b \right)}^{2}}=4a\left( x-c \right)$ is $x=c-a$
So the equation of directrix of parabola ${{y}^{2}}-4y-4x=0$ is $x=-2$
If we draw the graph we can visualize it more
We can see the black curve is parabola ${{y}^{2}}-4y-4x=0$ , the green line is the directrix $x=-2$ blue line is the axis. Point V is vertex and point F is the focus.
Note:
It is good to remember the formula for vertex, focus and directrix for standard parabola
We can observe that if we relocate the curve ${{y}^{2}}=4ax$ c units towards right and b units to upwards we will get the curve of${{\left( y-b \right)}^{2}}=4a\left( x-c \right)$. We can observe in the graph.
Location of focus is a unit from the vertex towards the inside of parabola along the axis. Directrix lies outside of parabola.
Note:
It is good to remember the formula for vertex, focus and directrix for standard parabola
We can observe that if we relocate the curve ${{y}^{2}}=4ax$ c units towards right and b units to upwards we will get the curve of${{\left( y-b \right)}^{2}}=4a\left( x-c \right)$. We can observe in the graph.
Location of focus is a unit from the vertex towards the inside of parabola along the axis. Directrix lies outside of parabola.
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