Parabola
3.0 Standard equation of Parabola
3.0 Standard equation of Parabola
1. ${y^2} = 4ax$
- Axis of parabola: $X$-axis i.e., $y=0$.
- Vertex $(0,0)$
- Focus $(a,0)$
- Equation of Directrix: $x+a=0$
- Equation of Latus rectum: $x-a=0$
- Point of intersection of parabola with latus rectum $(a, \pm 2a)$
- Length of latus rectum $ = \left| {4a} \right|$ or $4 \times $ distance between vertex and focus or $4 \times $ distance between vertex and directrix or $2 \times $ distance between focus and directrix .
2. ${x^2} = 4ay$
- Axis of parabola: $Y$-axis i.e., $x=0$.
- Vertex $(0,0)$
- Focus $(0,a)$
- Equation of Directrix: $y+a=0$
- Equation of Latus rectum: $y-a=0$
- Point of intersection of parabola with latus rectum $( \pm 2a,a)$
- Length of latus rectum $ = \left| {4a} \right|$ or $4 \times $ distance between vertex and focus or $4 \times $ distance between vertex and directrix or $2 \times $ distance between focus and directrix.
3. ${y^2} = -4ax$
- Axis of parabola: $X$-axis i.e., $y=0$.
- Vertex $(0,0)$
- Focus $(-a,0)$
- Equation of Directrix: $x-a=0$
- Equation of Latus rectum: $x+a=0$
- Point of intersection of parabola with latus rectum $(-a, \pm 2a)$
- Length of latus rectum $ = \left| {4a} \right|$ or $4 \times $ distance between vertex and focus or $4 \times $ distance between vertex and directrix or $2 \times $ distance between focus and directrix.
4. ${x^2} = -4ay$
- Axis of parabola: $Y$-axis i.e., $x=0$.
- Vertex $(0,0)$
- Focus $(0,-a)$
- Equation of Directrix: $y-a=0$
- Equation of Latus rectum: $y+a=0$
- Point of intersection of parabola with latus rectum $( \pm 2a,-a)$
- Length of latus rectum $ = \left| {4a} \right|$ or $4 \times $ distance between vertex and focus or $4 \times $ distance between vertex and directrix or $2 \times $ distance between focus and directrix.
Question 4. The equation of parabola is ${y^2} - 4y + 8 - 4x = 0$. Find the axis, vertex, focus, equation of directrix and latus rectum of parabola.
Solution: We can write the equation of parabola as $${\left( {y - 2} \right)^2} + 4 - 4x = 0$$ $${\left( {y - 2} \right)^2} = 4\left( {x - 1} \right)...(1)$$
Let us assume $Y = y - 2$ and $X = x - 1$, so equation $(1)$ becomes $${Y^2} = 4X...(2)$$
Compare the equation $(2)$ with the standard form of parabola i.e., ${y^2} = 4ax$ we get, $$a = 1$$
The axis of parabola given in equation $(2)$ is $$Y = 0$$
Therefore, the axis of parabola given in equation $(1)$ $$y-2=0$$
The vertex of parabola given in equation $(2)$ is $(0,0)$. Therefore, the vertex of parabola given in equation $(1)$ can be find out by putting $X=0$ and $Y=0$ i.e., $$x-1=0$$ and $$y-2=0$$
So, the vertex of parabola given in equation $(1)$ is $V(1,2)$.
The focus of parabola given in equation $(2)$ is $(a,0)$ i.e., $(1,0)$ and the focus of parabola given in equation $(1)$ can be find out by putting $X=a=1$ and $Y=0$ i.e., $$x-1=1$$ or, $$x=2$$ And $$y-2=0$$ Or, $$y=2$$
Therefore, the focus of parabola given in equation $(1)$ is $(2,2)$.
Equation of directrix of parabola given in equation $(1)$ is $$X+a=0$$
Or, $$x-1+1=0$$ $$x=0$$
Latus rectum of parabola given in equation $(1)$ is $$X-a=0$$ or, $$x-1-1=0$$ $$x=2$$
