Transformations of Conics Pure Math 30 If the

If the hyperbola x 2 - y 2 = -1 is stretched horizontally by

Given state the conic and its horizontal and vertical stretches. Remove the coefficients of

State the transformations when the equation y = x 2 becomes Solution: n Vertical

Given the ellipse , determine the new equation after a translation 3 units up

The ellipse is stretched horizontally by a factor of ½ and vertically by a

A tunnel has a semi-elliptical cross section. The maximum height of the tunnel is

Solution cont. : n We can see the horizontal stretch is 12. 5 and

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Transformations of Conics Pure Math 30

If the hyperbola x 2 - y 2 = -1 is stretched horizontally by a factor of 3 and vertically by a factor of ½, find the new equation. Solution: n First convert equation to standard form by dividing by – 1. x 2 - y 2 = -1 becomes -x 2 + y 2 = 1 n Now apply the stretches. 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.

Given state the conic and its horizontal and vertical stretches. Remove the coefficients of each variable and take the square root n Reciprocate the square root of the coefficient and you have the stretches n n Solution: The conic is an ellipse. Horizontal coefficient is 4/25. Square root is 2/5. Horizontal stretch is 5/2. n Vertical coefficient is 49/9. Square root is 7/3. Vertical stretch is 3/7. n

State the transformations when the equation y = x 2 becomes Solution: n Vertical stretch by a factor of 4 n Translations down. 2 units right and 2 units

Given the ellipse , determine the new equation after a translation 3 units up and 7 units right. Solution: n Determine original center point (2, -4) n Apply translations to this point (2 + 7, -4 + 3) n The new center is (9, -1) Put this back into equation.

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. n 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 n New equation becomes n

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? n Solution: Draw a diagram. (8. 5, ? ) 5. 5 m truck 12. 5 m 4 m

Solution cont. : n We can see the horizontal stretch is 12. 5 and the vertical stretch is 5. 5 to create the equation n Now sub in the given value of x (8. 5) to calculate y. Since the height of the truck is taller than the tunnel the truck will not fit.