# Coordinate Geometry Parabola

It is unlikely that you will get a problem on Coordinate geometry parabola on the new GRE. Its an uncommon topic but you are supposed to go through our article on Coordinate geometry parabola. Our article is definitely helpful in tackling problem on coordinate geometry parabola.

A parabola is the set of all points in a plane that are equidistant  from a fixed line & a fixed point (not on the line) in the plane.

The fixed line is called the directrix of the parabola & the fixed point F is called the focus.

$${ B }_{ 1 }{ P }_{ 1 }={ B }_{ 2 }{ P }_{ 2 }$$

A line through the focus & perpendicular to the directrix is called the axis of the parabola. The point of intersection of parabola with the axis is called the vertex of the parabola.

### Standard equations of parabola:

The equation of a parabola is simplest if the vertex is at the origin and the axis of symmetry is along the x-axis or y-axis. The four possible such orientation of parabola are shown in fig (a), fig (b), fig (c) & fig (d).

Fig(a)

Fig (b)

Fig (c)

Fig (d)

Let F be the focus & l be the directrix. In fig(a), the equation for the parabola with focus at (a,0) (where a>0) & directrix at x = a is

$${ y }^{ 2 }=4ax$$

In fig (b), the equation for the parabola with focus at (-a,0)  & directrix at x = +a  is

$${ y }^{ 2 }=-4ax$$

In fig (c), the equation for the parabola with focus at (0,a) & directrix at y = -a  is

$${ x }^{ 2 }=4ay$$

In fig (d), the equation for the parabola with focus at (0,-a) & directrix at y = a is

$${ y }^{ 2 }=4ax$$…… (1)

$${ y }^{ 2 }=-4ax$$…… (2)

$${ x }^{ 2 }=4ay$$…… (3)

$${ x }^{ 2 }=-4ay$$…… (4)

These four equations are known as standard equation of the parabola, derivation is beyond the scope here.

For any positive number c,

• The graph of parabola $${ y }^{ 2 }=4ax$$ when shifts upward by c units, the equation of parabola becomes
$${ (y-c) }^{ 2 }=4ax$$
• The graph of parabola $${ y }^{ 2 }=4ax$$ when shifts downward by c units, the equation of parabola becomes
$${ (y+c) }^{ 2 }=4ax$$
• The equation of parabola $${ y }^{ 2 }=4ax$$ when shifts to the left by c units, the equation of parabola becomes
$${ y }^{ 2 }=4a(x+c)$$
• The equation of parabola $${ y }^{ 2 }=4ax$$ when shift to the right by c units, the equation of parabola becomes
$${ y }^{ 2 }=4a(x-c)$$

Similar rules apply to all the standard parabolas.

Problem   The equation $$y={ x }^{ 2 }-2x-3$$ has the following graph. Find the following:

• Coordinates of the vertex
• The y-intercept

Solution:  The equation

$$y={ x }^{ 2 }-2x-3\\ y={ x }^{ 2 }-2x+1-4\\ (y+4)={ (x-1) }^{ 2 }\\ { (x-1) }^{ 2 }=4(\frac { 1 }{ 4 } )(y+4)$$

Therefore, the point (1,-4) is the vertex of parabola.

The y-intercept is the y-coordinate of the point on the parabola at which  x=0, which is

$$y=0-2(0)-3={ (-3) }^{ Ans }$$

Problem

Which of the following could be the equation of the parabola in the coordinate plane above?

a) $$y={ x }^{ 2 }+3$$

b) $$y={ (x-3) }^{ 2 }+3$$

c) $$y={ (x+3) }^{ 2 }-3$$

d) $$y={ (x-3) }^{ 2 }-3$$

e) $$y={ (x+3) }^{ 2 }+3$$

Sol.

This graph has been obtained by shifting standard parabola (having vertex at (0,0)  & focus at (0,a)  k units upward & h units toward the left.

$${ x }^{ 2 }=4ay$$—— (equation of standard parabola)

Equation after shifting vertex at (-h,k)

$${ (x+h) }^{ 2 }=4a(y-k)\quad \quad (h>0,K>0)\\ { (x+h) }^{ 2 }+4ak=4ay$$

After analysing answer choices, we can easily reckon that choice $$y={ (x+3) }^{ 2 }+3$$ places the vertex in the correct quadrant.

Problem

Which of the following could be the equation of the parabola in the coordinate place above?

a) $$y=-x-1$$

b) $$y={ x }^{ 2 }+1$$

c) $$y={ -x }^{ 2 }-1$$

d) $$y={ -x }^{ 2 }+1$$

e) $$y=-{ (x-1) }^{ 2 }$$

Sol.   This graph has been obtained by shifting standard parabola (having vertex at (0,0) & focus at (0, –a) 1 unit upward.

$${ x }^{ 2 }=-4ay——-$$(Equation of parabola having vertex at (0,0) & focus at (0, –a)).

Equation after shifting vertex at (0,1):

$${ x }^{ 2 }=-4a(y-1)\quad \quad (a>0,\quad let\quad a=\frac { 1 }{ 4 } )\\ { x }^{ 2 }=-4(\frac { 1 }{ 4 } )(y-1)\\ { x }^{ 2 }=-(y-1)=-y+1\\ ({ y=-{ x }^{ 2 }+1) }^{ Ans }$$

Problem

If the equation of the parabola in the coordinate plane above is $$y={ (x-h) }^{ 2 }+k$$ and (-3,n) is a point on the parabola, what is the value of n?

Sol.   This graph has been obtained by shifting standard parabola (having vertex at (0,0)  & focus at (0,a)) 2 units rightward.

$${ x }^{ 2 }=4ay$$— — (Equation of standard parabola having vertex at (0,0) and focus at (0,a)).

$${ (x-2) }^{ 2 }=4ay$$— — (Equation of standard parabola by shifting vertex at (2,0)).

$${ (x-2) }^{ 2 }=4ay$$

Let’s compare this equation with given equation

$$y={ (x-h) }^{ 2 }+k\\ 4ay={ (x-2) }^{ 2 }+k\\ Therefore,\quad a=\frac { 1 }{ 4 } ,\quad h=2,\quad k=0\\ y={ (x-2) }^{ 2 }$$

Point (-3,n) lies on this parabola

$$n={ (-3-2) }^{ 2 }\\ n=25\\ Therefore,\quad { (n=25) }^{ Ans }$$

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