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Worksheet: Equation of a Parabola

Q1:

The figure shows the parabola π‘₯ = 2 𝑦 βˆ’ 1 6 𝑦 + 2 2 2 with its vertex 𝑉 marked.

What are the coordinates of 𝑉 ?

  • A ( βˆ’ 4 , 1 0 )
  • B ( βˆ’ 6 , 4 )
  • C ( βˆ’ 4 , 6 )
  • D ( βˆ’ 1 0 , 4 )
  • E ( 4 , βˆ’ 1 0 )

Q2:

Find the equation of a parabola with a focus of ( βˆ’ 1 , βˆ’ 3 ) and a directrix of 𝑦 = βˆ’ 5 . Give your answer in the form 𝑦 = π‘Ž π‘₯ + 𝑏 π‘₯ + 𝑐 2 .

  • A 𝑦 = 1 2 π‘₯ βˆ’ 1 4 π‘₯ βˆ’ 1 5 4 2
  • B 𝑦 = 1 2 π‘₯ + 1 4 π‘₯ + 1 5 4 2
  • C 𝑦 = 1 4 π‘₯ + 1 2 π‘₯ + 1 5 4 2
  • D 𝑦 = 1 4 π‘₯ + 1 2 π‘₯ βˆ’ 1 5 4 2
  • E 𝑦 = 1 2 π‘₯ + π‘₯ βˆ’ 1 5 4 2

Q3:

Complete the following definition: A parabola is defined as the set of all points a fixed point called the focus and a fixed line called the directrix.

  • Acentered between
  • Bat a given distance from
  • Cwith a radius from
  • Dequidistant from
  • Ewith a diameter from

Q4:

The diagram shows a parabola with a horizontal axis whose vertex is ( β„Ž , π‘˜ ) . The focus 𝐹 , directrix 𝑑 , and a point ( π‘₯ , 𝑦 ) on the parabola are marked.

The distance from the vertex to the focus is equal to the distance from the vertex to the directrix. Let this distance be 𝑝 .

Write the coordinates of the focus in terms of β„Ž , 𝑝 , and π‘˜ .

  • A ( π‘˜ , β„Ž + 𝑝 )
  • B ( β„Ž βˆ’ 𝑝 , π‘˜ )
  • C ( π‘˜ , β„Ž βˆ’ 𝑝 )
  • D ( β„Ž + 𝑝 , π‘˜ )
  • E ( β„Ž + 𝑝 , βˆ’ π‘˜ )

Write an expression for the distance from the point ( π‘₯ , 𝑦 ) to the focus.

  • A  ( π‘₯ βˆ’ ( β„Ž + 𝑝 ) ) + ( 𝑦 βˆ’ π‘˜ ) 2 2
  • B  ( π‘₯ βˆ’ π‘˜ ) + ( 𝑦 βˆ’ ( β„Ž βˆ’ 𝑝 ) ) 2 2
  • C  ( π‘₯ βˆ’ π‘˜ ) + ( 𝑦 βˆ’ ( β„Ž + 𝑝 ) ) 2 2
  • D  ( π‘₯ βˆ’ ( β„Ž βˆ’ 𝑝 ) ) + ( 𝑦 βˆ’ π‘˜ ) 2 2
  • E  ( π‘₯ βˆ’ ( β„Ž + 𝑝 ) ) + ( 𝑦 + π‘˜ ) 2 2

Write an equation for the directrix.

  • A π‘₯ = βˆ’ β„Ž 𝑝
  • B π‘₯ = β„Ž + 𝑝
  • C π‘₯ = β„Ž βˆ’ 𝑝
  • D π‘₯ = β„Ž 𝑝
  • E π‘₯ = 𝑝 βˆ’ β„Ž

Write an expression for the distance between the point ( π‘₯ , 𝑦 ) and the directrix.

  • A π‘₯ βˆ’ ( β„Ž βˆ’ 𝑝 )
  • B π‘₯ + ( β„Ž + 𝑝 )
  • C π‘₯ + ( β„Ž βˆ’ 𝑝 )
  • D π‘₯ βˆ’ ( β„Ž + 𝑝 )
  • E π‘₯ βˆ’ ( β„Ž 𝑝 )

A parabola can be defined as the locus of points that are equidistant from a fixed line (the directrix) and a fixed point that is not on the line (the focus).

By equating your expressions, squaring both sides, and rearranging, write an equation for ( 𝑦 βˆ’ π‘˜ ) 2 in terms of π‘₯ , 𝑝 , and β„Ž that describes the parabola.

  • A ( 𝑦 + π‘˜ ) = 𝑝 ( π‘₯ + β„Ž ) 2
  • B ( 𝑦 + π‘˜ ) = 4 𝑝 ( π‘₯ + β„Ž ) 2
  • C ( 𝑦 βˆ’ π‘˜ ) = 𝑝 ( π‘₯ βˆ’ β„Ž ) 2
  • D ( 𝑦 βˆ’ π‘˜ ) = 4 𝑝 ( π‘₯ βˆ’ β„Ž ) 2
  • E ( 𝑦 βˆ’ β„Ž ) = 4 𝑝 ( π‘₯ βˆ’ π‘˜ ) 2

Q5:

An arch in the shape of a parabola has a span of 160 feet and a maximum height of 40 feet. Find the equation of the parabola. At what distance from the center is the arch 20 ft high?

  • A π‘₯ = 1 6 0 ( 𝑦 + 4 0 ) , π‘₯ = 4 0 √ 6 2 ft
  • B π‘₯ = 1 6 0 𝑦 , π‘₯ = 4 0 √ 2 2 ft
  • C 𝑦 = βˆ’ 1 6 0 ( π‘₯ βˆ’ 4 0 ) , π‘₯ = 3 7 . 5 2 ft
  • D π‘₯ = βˆ’ 1 6 0 ( 𝑦 βˆ’ 4 0 ) , π‘₯ = 4 0 √ 2 2 ft
  • E π‘₯ = βˆ’ 4 ( 4 0 𝑦 βˆ’ 4 0 ) , π‘₯ = 4 √ 2 1 0 2 ft

Q6:

A parabola has the equation π‘₯ = 3 2 𝑦 2 .

What are the coordinates of its focus?

  • A ( 0 , 6 )
  • B ο€Ό 3 8 , 0 
  • C ( 6 , 0 )
  • D ο€Ό 0 , 3 8 
  • E ο€Ό 0 , 3 2 

Write an equation for its directrix.

  • A 𝑦 + 3 8 = 0
  • B 𝑦 βˆ’ 6 = 0
  • C 𝑦 + 6 = 0
  • D 𝑦 βˆ’ 3 8 = 0
  • E 𝑦 + 3 2 = 0

Q7:

Find an equation for the parabola whose focus is the point ( βˆ’ 5 , βˆ’ 1 ) and whose directrix is the line 𝑦 + 1 2 = 0 .

  • A ( π‘₯ + 5 ) = 1 4 ( 𝑦 + 1 ) 2 2
  • B ( π‘₯ βˆ’ 5 ) = 2 2 ( 2 𝑦 βˆ’ 1 ) 2 2
  • C ( π‘₯ βˆ’ 5 ) = 1 4 ( 𝑦 βˆ’ 1 ) 2 2
  • D ( π‘₯ + 5 ) = 2 2 ( 𝑦 + 1 ) 2 2
  • E ( π‘₯ + 5 ) = 1 2 ( 𝑦 + 1 ) 2 2

Q8:

Find the equation of the parabola with focus ( 3 , βˆ’ 2 ) and directrix 𝑦 = βˆ’ 3 2 . Give your answer in the form 𝑦 = π‘Ž π‘₯ + 𝑏 π‘₯ + 𝑐 2 .

  • A 𝑦 = π‘₯ + 6 π‘₯ + 4 3 4 2
  • B 𝑦 = βˆ’ π‘₯ + 6 π‘₯ + 4 3 4 2
  • C 𝑦 = βˆ’ π‘₯ + 6 π‘₯ βˆ’ 2 1 2 2
  • D 𝑦 = βˆ’ π‘₯ + 6 π‘₯ βˆ’ 4 3 3 2
  • E 𝑦 = π‘₯ + 6 π‘₯ βˆ’ 2 1 2 2

Q9:

The mirror in an automobile headlight has a parabolic cross section with the light bulb at the focus. On a schematic, the equation of the parabola is given as π‘₯ = 4 𝑦 2 . At what coordinates should you place the light bulb?

  • A ( 1 , 0 )
  • B ( 0 , βˆ’ 1 )
  • C ( βˆ’ 1 , 0 )
  • D ( 0 , 1 )
  • E ( 2 , 1 )

Q10:

The diagram shows a parabola that is symmetrical about the 𝑦 -axis and whose vertex is at the origin. Its Cartesian equation is π‘₯ = 4 𝑝 𝑦 2 , where 𝑝 is a positive constant. The focus of the parabola is the point ( 0 , 𝑝 ) and the directrix is the line with equation 𝑦 = βˆ’ 𝑝 .

Find the Cartesian equation of the parabola whose focus is the point ο€Ό 0 , 5 4  and whose directrix is the line 𝑦 = βˆ’ 5 4 .

  • A π‘₯ = 5 𝑦 2
  • B 𝑦 = 5 4 π‘₯ 2
  • C π‘₯ = 2 0 𝑦 2
  • D π‘₯ = 5 𝑦 2
  • E π‘₯ = 2 0 𝑦 2

Q11:

Consider the parabola whose vertex is the point ( βˆ’ 5 , 4 ) and whose directrix is the line π‘₯ = βˆ’ 1 .

What is the distance from the vertex to the directrix?

Find an equation for the parabola.

  • A ( 𝑦 βˆ’ 4 ) = βˆ’ 1 6 ( π‘₯ + 5 ) 2
  • B ( 𝑦 βˆ’ 4 ) = βˆ’ 2 0 ( π‘₯ + 5 ) 2
  • C ( 𝑦 + 4 ) = βˆ’ 1 6 ( π‘₯ βˆ’ 5 ) 2
  • D ( 𝑦 βˆ’ 4 ) = βˆ’ 4 ( π‘₯ + 5 ) 2
  • E ( 𝑦 + 4 ) = βˆ’ 2 0 ( π‘₯ βˆ’ 5 ) 2

Q12:

Consider the graph:

Which of the following could be the equation of the parabola?

  • A 𝑦 = ( π‘₯ βˆ’ 1 ) ( π‘₯ βˆ’ 5 )
  • B 𝑦 = ( π‘₯ + 1 ) ( π‘₯ + 5 )
  • C 𝑦 = βˆ’ ( π‘₯ βˆ’ 1 ) ( π‘₯ βˆ’ 5 )
  • D 𝑦 = βˆ’ ( π‘₯ + 1 ) ( π‘₯ + 5 )
  • E 𝑦 = βˆ’ ( π‘₯ + 1 ) ( π‘₯ βˆ’ 5 )

Q13:

The given figure shows a parabola with a focus of ( π‘Ž , 𝑏 ) , a directrix at 𝑦 = π‘˜ , and a general point ( π‘₯ , 𝑦 ) .

Find an expression for the length of the line from ( π‘₯ , 𝑦 ) to the focus.

  • A √ ( π‘₯ + π‘Ž ) + ( 𝑦 + 𝑏 ) 2 2
  • B √ ( π‘₯ βˆ’ π‘Ž ) βˆ’ ( 𝑦 βˆ’ 𝑏 ) 2 2
  • C √ ( π‘₯ βˆ’ π‘Ž ) + ( 𝑦 βˆ’ 𝑏 )
  • D √ ( π‘₯ βˆ’ π‘Ž ) + ( 𝑦 βˆ’ 𝑏 ) 2 2
  • E √ ( π‘₯ + π‘Ž ) + ( 𝑦 + 𝑏 )

Write an expression for the distance between ( π‘₯ , 𝑦 ) and the directrix 𝑦 = π‘˜ .

  • A 𝑦 βˆ’ π‘˜
  • B π‘₯ + π‘˜
  • C π‘₯ βˆ’ π‘˜
  • D 𝑦 + π‘˜
  • E ( 𝑦 βˆ’ π‘˜ ) 2

Equate the two expressions and square both sides.

  • A ( π‘₯ βˆ’ 𝑏 ) βˆ’ ( 𝑦 βˆ’ π‘Ž ) = ( 𝑦 + π‘˜ ) 2 2 2
  • B ( π‘₯ βˆ’ π‘Ž ) + ( 𝑦 βˆ’ 𝑏 ) = ( 𝑦 + π‘˜ ) 2 2 2
  • C ( π‘₯ βˆ’ π‘Ž ) + ( 𝑦 βˆ’ 𝑏 ) = ( 𝑦 βˆ’ π‘˜ ) 2 2 2
  • D ( π‘₯ βˆ’ 𝑏 ) + ( 𝑦 βˆ’ π‘Ž ) = ( 𝑦 βˆ’ π‘˜ ) 2 2 2
  • E ( π‘₯ βˆ’ π‘Ž ) βˆ’ ( 𝑦 βˆ’ 𝑏 ) = ( 𝑦 βˆ’ π‘˜ ) 2 2 2

Expand and simplify the expressions excluding ( π‘₯ βˆ’ π‘Ž ) 2 , and then make 𝑦 the subject and simplify.

  • A 𝑦 = 1 2 ο€Ύ ( π‘₯ βˆ’ π‘Ž ) 𝑏 βˆ’ π‘˜ + 𝑏 + π‘˜  2
  • B 𝑦 = ο€Ύ ( π‘₯ βˆ’ π‘Ž ) 𝑏 + π‘˜ βˆ’ 𝑏 + π‘˜  2
  • C 𝑦 = 1 2 ο€Ύ ( π‘₯ βˆ’ π‘Ž ) 𝑏 βˆ’ π‘˜ βˆ’ 𝑏 + π‘˜  2
  • D 𝑦 = 1 2 ο€Ύ ( π‘₯ βˆ’ π‘Ž ) 𝑏 βˆ’ π‘˜ + 𝑏 βˆ’ π‘˜  2
  • E 𝑦 = 1 2 ο€Ύ ( π‘₯ βˆ’ π‘Ž ) 𝑏 + π‘˜ + 𝑏 βˆ’ π‘˜  2

Q14:

Find the focus and directrix of the parabola 𝑦 = 2 π‘₯ + 5 π‘₯ + 4 2 .

  • A focus: ο€Ό βˆ’ 4 5 , 1  , directrix: 𝑦 = 3 4
  • B focus: ο€Ό 5 4 , 1  , directrix: 𝑦 = βˆ’ 4 3
  • C focus: ο€Ό 4 4 , 1  , directrix: 𝑦 = 3 4
  • D focus: ο€Ό βˆ’ 5 4 , 1  , directrix: 𝑦 = 3 4
  • E focus: ο€Ό βˆ’ 5 4 , 1  , directrix: 𝑦 = βˆ’ 4 3

Q15:

A parabola has the equation 𝑦 = √ 1 0 3 π‘₯ 2 .

What are the coordinates of its focus?

  • A ο€Ώ 4 √ 1 0 3 , 0 
  • B ο€Ώ 0 , √ 1 0 1 2 
  • C ο€Ώ βˆ’ √ 1 0 1 2 , 0 
  • D ο€Ώ √ 1 0 1 2 , 0 
  • E ο€Ώ √ 1 0 3 , 0 

Write an equation for its directrix.

  • A π‘₯ = βˆ’ √ 1 0 1 2
  • B π‘₯ = √ 1 0 3
  • C π‘₯ = 4 √ 1 0 3
  • D π‘₯ = √ 1 0 1 2
  • E 𝑦 = βˆ’ √ 1 0 1 2

Q16:

A parabola has the equation π‘₯ = 2 √ 2 𝑦 2 .

What are the coordinates of its focus?

  • A ο€» 0 , 8 √ 2 
  • B ο€Ώ √ 2 2 , 0 
  • C ο€» 8 √ 2 , 0 
  • D ο€Ώ 0 , √ 2 2 
  • E ο€» 0 , 2 √ 2 

Write an equation for its directrix.

  • A 𝑦 + √ 2 2 = 0
  • B 𝑦 βˆ’ 8 √ 2 = 0
  • C 𝑦 + 8 √ 2 = 0
  • D 𝑦 βˆ’ √ 2 2 = 0
  • E 𝑦 + 2 √ 2 = 0

Q17:

A parabola has equation ( 𝑦 + 2 ) = 1 2 ( π‘₯ βˆ’ 1 ) 2 .

Find the coordinates of the vertex.

  • A ( 7 , βˆ’ 2 )
  • B ( βˆ’ 1 , 2 )
  • C ( βˆ’ 7 , 2 )
  • D ( 1 , βˆ’ 2 )
  • E ( βˆ’ 2 , 1 )

Determine the equation of the directrix.

  • A π‘₯ + 2 = 0
  • B π‘₯ βˆ’ 7 = 0
  • C π‘₯ + 7 = 0
  • D π‘₯ βˆ’ 2 = 0
  • E π‘₯ + 1 = 0

Q18:

Write an equation for the parabola whose focus is the point ( βˆ’ 4 , βˆ’ 3 ) and whose directrix is the line π‘₯ = 0 .

  • A ( 𝑦 βˆ’ 3 ) = βˆ’ 8 ( π‘₯ βˆ’ 2 ) 2
  • B ( 𝑦 + 3 ) = βˆ’ 8 ( π‘₯ + 4 ) 2
  • C ( 𝑦 βˆ’ 3 ) = βˆ’ 8 ( π‘₯ βˆ’ 4 ) 2
  • D ( 𝑦 + 3 ) = βˆ’ 8 ( π‘₯ + 2 ) 2
  • E ( 𝑦 + 3 ) = βˆ’ 2 ( π‘₯ + 2 ) 2

Q19:

A satellite dish is shaped like a paraboloid of revolution. This means that it can be formed by rotating a parabola around its axis of symmetry. The receiver is to be located at the focus. If the dish is 8 feet across at its opening and 2 feet deep, how far should the receiver be placed above the vertex?

Q20:

A satellite dish is shaped like a paraboloid of revolution. This means that it can be formed by rotating a parabola around its axis of symmetry. The receiver is to be located at the focus. If the dish is 12 feet across at its opening and 4 feet deep at its center, how far should the receiver be placed above the vertex?

Q21:

The diagram shows a parabola that is symmetrical about the π‘₯ -axis and whose vertex is at the origin. Its Cartesian equation is 𝑦 = 4 𝑝 π‘₯ 2 , where 𝑝 is a positive constant. The focus of the parabola is the point ( 𝑝 , 0 ) and the directrix is the line with equation π‘₯ + 𝑝 = 0 .

Find the Cartesian equation of the parabola whose focus is the point ο€Ό 3 2 , 0  and whose directrix is the line π‘₯ + 3 2 = 0 .

  • A 𝑦 = 6 π‘₯ 2
  • B 𝑦 = 3 2 π‘₯ 2
  • C 𝑦 = 1 2 π‘₯ 2
  • D 𝑦 = 6 π‘₯ 2
  • E 𝑦 = 1 2 π‘₯ 2

Q22:

The given figure shows a parabola with a focus of (3, 2), a directrix at 𝑦 = 1 , and a general point ( π‘₯ , 𝑦 ) .

Find an expression for the length of the line from the point ( π‘₯ , 𝑦 ) to the point (3, 2).

  • A √ ( π‘₯ βˆ’ 2 ) + ( 𝑦 βˆ’ 3 )  
  • B √ ( π‘₯ βˆ’ 3 ) βˆ’ ( 𝑦 βˆ’ 2 )  
  • C √ ( π‘₯ βˆ’ 2 ) βˆ’ ( 𝑦 βˆ’ 3 )  
  • D √ ( π‘₯ βˆ’ 3 ) + ( 𝑦 βˆ’ 2 )  
  • E √ ( π‘₯ βˆ’ 3 ) + ( 𝑦 βˆ’ 2 )

Write an expression for the distance between ( π‘₯ , 𝑦 ) and the directrix 𝑦 = 1 .

  • A 𝑦 βˆ’ 1
  • B √ π‘₯ + ( 𝑦 βˆ’ 1 )  
  • C π‘₯ βˆ’ 1
  • D √ π‘₯ βˆ’ ( 𝑦 βˆ’ 1 )  
  • E ( 𝑦 βˆ’ 1 ) 

By equating the two expressions from ( π‘Ž ) and ( 𝑏 ) , work out an equation for the parabola. Give your answer in the form 𝑦 = π‘Ž π‘₯ + 𝑏 π‘₯ + 𝑐  .

  • A 𝑦 = 1 6 π‘₯ βˆ’ π‘₯ + 2 
  • B 𝑦 = 1 2 π‘₯ βˆ’ 3 π‘₯ + 7 
  • C 𝑦 = 1 2 π‘₯ βˆ’ 3 π‘₯ + 6 
  • D 𝑦 = 1 2 π‘₯ + 3 π‘₯ + 6 
  • E 𝑦 = 1 2 π‘₯ + 3 π‘₯ + 7 

Q23:

The diagram shows a parabola that is symmetrical about the π‘₯ -axis and whose vertex is at the origin. Its Cartesian equation is 𝑦 = βˆ’ 4 𝑝 π‘₯ 2 , where 𝑝 is a positive constant. The focus of the parabola is the point ( βˆ’ 𝑝 , 0 ) and the directrix is the line with equation π‘₯ = 𝑝 .

Consider the parabola with Cartesian equation 𝑦 = βˆ’ 1 4 π‘₯ 2 .

What are the coordinates of the focus of the parabola with Cartesian equation 𝑦 = βˆ’ 1 4 π‘₯ 2 ?

  • A ο€Ό 7 2 , 0 
  • B ο€Ό 0 , βˆ’ 7 2 
  • C ο€Ό 0 , 7 2 
  • D ο€Ό βˆ’ 7 2 , 0 
  • E ( βˆ’ 1 4 , 0 )

Write the equation of its directrix.

  • A π‘₯ = 7 2
  • B π‘₯ = 1 4
  • C π‘₯ = βˆ’ 1 4
  • D π‘₯ = βˆ’ 7 2
  • E π‘₯ = 7

Q24:

An arch is in the shape of a parabola. It has a span of 100 feet and a maximum height of 20 feet. Find the equation of the parabola, and determine the height of the arch 40 feet from the center.

  • A π‘₯ = 4 ( 3 1 . 2 5 𝑦 βˆ’ 2 0 ) 2 , β„Ž = 1 3 . 4 4 f t
  • B π‘₯ = 1 2 5 𝑦 2 , β„Ž = 1 2 . 8 f t
  • C π‘₯ = 1 2 5 𝑦 βˆ’ 2 0 2 , β„Ž = 1 2 . 9 6 f t
  • D π‘₯ = βˆ’ 1 2 5 ( 𝑦 βˆ’ 2 0 ) 2 , β„Ž = 7 . 2 f t
  • E π‘₯ = βˆ’ 1 2 5 𝑦 βˆ’ 2 0 2 , β„Ž = 7 . 2 f t

Q25:

Write an equation for the parabola whose focus is the point ο€Ώ 0 , βˆ’ √ 2 3  and whose directrix is the line 𝑦 = √ 2 3 .

  • A π‘₯ = √ 2 1 2 𝑦 2
  • B π‘₯ = √ 2 3 𝑦 2
  • C 𝑦 = 4 √ 2 3 π‘₯ 2
  • D π‘₯ = βˆ’ 4 √ 2 3 𝑦 2
  • E 𝑦 = √ 2 1 2 π‘₯ 2