Transformations of
Conics
Pure Math 30
If the hyperbola x2 - y2 = -1 is stretched
horizontally by a factor of 3 and vertically by a
factor of ½, find the new equation.
Solution:

First convert equation to standard form by
dividing by –1.

x2 - y2 = -1 becomes -x2 + y2 = 1
Now apply the stretches.
x
y

1
2
2
(3) ( 12 )
2
2
x
y

1
1
9
4
2
2
So the equation becomes
Because fractions in the
denominator look incorrect,
we convert by remembering
that dividing by a fraction is
the same as multiplying by
the reciprocal.
 x2
 4 y2  1
9
2
2
4(
x

2)
49(
y

3)
Given

 1, state the conic
25
9
and its horizontal and vertical stretches.
Remove the coefficients of each variable and
take the square root
 Reciprocate the square root of the coefficient
and you have the stretches


Solution:

Horizontal coefficient is 4/25. Square root is 2/5.
Horizontal stretch is 5/2.
Vertical coefficient is 49/9. Square root is 7/3.
Vertical stretch is 3/7.

The conic is an ellipse.
State the transformations when the equation y = x2
becomes
y  2  4( x  2)
2
Solution:
 Vertical stretch by a factor of 4

Translations 2 units right and 2 units
down.
Given the ellipse
( x  2) ( y  4)
,

1
9
16
2
2
determine the new equation after a translation
3 units up and 7 units right.
Solution:
 Determine original center point (2, -4)
 Apply translations to this point (2 + 7, -4 + 3)
 The new center is (9, -1) Put this back into
equation.
( x  9) ( y  1)

1
9
16
2
2
2
2
x
y

1
9 16
The ellipse
is stretched
horizontally by a factor of ½ and vertically by a
factor of 3. Determine the new equation.
Solution: Remove the stretches from the
equation. H.s. is 3 and v.s. is 4.
 Multiply by the new stretches and put these
values back into equation. H.s. becomes
3 x ½ = 3/2.
V.s. becomes 4 x 3 = 12
 New equation becomes

x
2
2
y

1
2
2
( 3 2) (12)
x
9
2
4
2
y

1
144
2
2
4x
y

1
9 144
A tunnel has a semi-elliptical cross section. The
maximum height of the tunnel is 5.5 m, and the
full tunnel width is 25 m. A truck in the right
lane is 4.3 m tall, and will be 4 m away from the
tunnel wall. Will the truck be able to get
through the tunnel?

Solution: Draw a diagram.
(8.5, ?)
5.5 m
truck
12.5 m
4m
Solution cont. :

We can see the horizontal stretch is 12.5 and the
vertical stretch is 5.5 to create the equation
x2
y2
 2 1
2
12.5 5.5

Now sub in the given value of x (8.5) to calculate y.
2
2
8.5
y

1
2
2
12.5 5.5
2
2
y
8.5
 1
2
5.5
12.52
2
y
 .5376
2
5.5
2
y  16.2624
y  4.03 m
Since the
height of the
truck is
taller than
the tunnel
the truck
will not fit.