The curve CC has equation x2−4y2=4x^2 - 4y^2 = 4.

  1. (a)

    Sketch CC, stating the equations of any asymptotes and the coordinates of any points of intersection with the axes.

  2. (b)

    Describe a pair of transformations which transforms the graph of CC onto the graph of (x−2)2−y2=4(x - 2)^2 - y^2 = 4.

Work through it with AI →

It won't just hand you the answer — it asks where you're stuck.How the checker works

Worked solution

5(a)

1. Rewrite the equation into the standard form of a hyperbola to identify its shape.

x2−4y2=4x^2 - 4y^2 = 4
Dividing by 44:
x24−y21=1\frac{x^2}{4} - \frac{y^2}{1} = 1
This represents a hyperbola centred at the origin, opening horizontally.

2. Find the points of intersection with the axes.

To find xx-intercepts, let y=0y = 0:
x24=1  ⟹  x2=4  ⟹  x=±2\frac{x^2}{4} = 1 \implies x^2 = 4 \implies x = \pm 2
The points of intersection with the xx-axis are (2,0)(2, 0) and (−2,0)(-2, 0).

To find yy-intercepts, let x=0x = 0:
−y21=1  ⟹  y2=−1-\frac{y^2}{1} = 1 \implies y^2 = -1
Since there are no real solutions, the curve does not intersect the yy-axis.

3. Determine the equations of the asymptotes.

For a hyperbola of the form x2a2−y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, the asymptotes are given by y=±baxy = \pm \frac{b}{a}x.
Here a=2a = 2 and b=1b = 1, so the asymptotes are:
y=±12xy = \pm \frac{1}{2}x

4. Sketch the hyperbola $C$.

  1. Draw the coordinate axes.
  2. Plot the xx-intercepts at (−2,0)(-2, 0) and (2,0)(2, 0).
  3. Draw and label the dashed lines for the asymptotes y=12xy = \frac{1}{2}x and y=−12xy = -\frac{1}{2}x.
  4. Sketch the two branches of the hyperbola opening to the left and right, passing through the intercepts and becoming asymptotic to the dashed lines as x→±∞x \to \pm\infty.
Answer

Asymptotes: y=12xy = \frac{1}{2}x and y=−12xy = -\frac{1}{2}x
Intersections with axes: (2,0)(2, 0) and (−2,0)(-2, 0)
[A sketch showing a horizontal hyperbola with branches in the left and right quadrants, passing through the intercepts and approaching the dashed asymptote lines.]

5(b)

1. Identify the algebraic substitutions that link the original equation to the new equation.

The new equation is (x−2)2−y2=4(x - 2)^2 - y^2 = 4.
This can be rewritten to match the form of the original equation CC:
(x−2)2−4(y2)2=4(x - 2)^2 - 4\left(\frac{y}{2}\right)^2 = 4
Comparing this with x2−4y2=4x^2 - 4y^2 = 4, we observe that xx is replaced by (x−2)(x - 2) and yy is replaced by y2\frac{y}{2}.

2. Describe the geometric transformations corresponding to these substitutions.

Replacing xx with (x−2)(x - 2) corresponds to a translation of 22 units in the positive xx-direction.
Replacing yy with y2\frac{y}{2} corresponds to a scaling (or stretch) parallel to the yy-axis with a scale factor of 22.
Since these transformations act on different, orthogonal axes, they can be applied in either order.

Answer
  1. Translation of 22 units in the positive xx-direction.
  2. Stretch parallel to the yy-axis with scale factor 22.

AI analysis — may contain mistakes.

Worked this one out on paper? Snap your steps and Gigle's AI math checker will circle the step where it went wrong.

More questions on this topic

  1. 1The curve CC has equation y=x2+x+3x−2y = \frac{x^2 + x + 3}{x - 2}. Without using a calculator, find the set of values that yy…
  2. 2The functions ff and gg are defined by f:x↦16−4x,  x∈R,  x≤4,f : x \mapsto \sqrt{16 - 4x},\; x \in \mathbb{R},\; x \le 4, g:x↦x2,  x∈R.g : x \mapsto x^{2},\; x \in \mathbb{R}.…
  3. 3By expressing the equation of the curve y=12x+112x+1y = \frac{12x + 11}{2x + 1} in the form y=A+B2x+1y = A + \frac{B}{2x + 1}, where AA…
  4. 45 (a) Solve the following integral. ∫1−2x3+2x−x2 dx.\displaystyle\int \frac{1-2x}{\sqrt{3+2x-x^2}}\,\mathrm{d}x. ∫xln⁡(2−x2) dx\displaystyle\int x\ln\left(2-x^2\right)\,\mathrm{d}x…
  5. 5A curve C1C_1 has the equation p2x2−y2=p2p^2x^2 - y^2 = p^2 where p>1p > 1. Sketch C1C_1, stating the coordinates of any points o…