<p>A curve $C$ has equation $y=\frac{1}{\sqrt{4ax-{{x}^{2}}}}$, where $a > 0$.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Sketch $C$ and give the equations of any asymptotes, in terms of $a$ where appropriate.</span></div><p style="text-align: right;">[4]</p></li><li><div class="list-item-content"><span class="list-item-text">Find the smallest possible value of $y$ in terms of $a$.</span></div><p style="text-align: right;">[1]</p></li><li><div class="list-item-content"><span class="list-item-text">Describe the transformation that maps the graph of $C$ onto the graph of $y=\frac{1}{\sqrt{4{{a}^{2}}-{{x}^{2}}}}$.</span></div><p style="text-align: right;">[3]</p></li></ol>

A curve $C$ has equation $y=\frac{1}{\sqrt{4ax-{{x}^{2}}}}$, where $a > 0$.

Sketch $C$ and give the equations of any asymptotes, in terms of $a$ where appropriate.

Find the smallest possible value of $y$ in terms of $a$.

Describe the transformation that maps the graph of $C$ onto the graph of $y=\frac{1}{\sqrt{4{{a}^{2}}-{{x}^{2}}}}$.

Transformations of Graphs - General

Transformations of Graphs

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

N2021 P1 Q6

$y = \frac{1}{\sqrt{4ax - x^2}} = \frac{1}{\sqrt{x(4a-x)}}$

Min of $y$ occurs when $\sqrt{4a^2 - (x-2a)^2}$ is maximum

$y = \frac{1}{\sqrt{4a^2}} = \frac{1}{2a}$

[Graph sketch with vertical asymptotes at $x = 0$ and $x = 4a$, minimum point at $(2a, \frac{1}{2a})$]

$\therefore$ minimum value of $y = \frac{1}{2a}$

$y = \frac{1}{\sqrt{4a^2 - (x-2a)^2}} \rightarrow y = \frac{1}{\sqrt{4a^2 - x^2}}$

$\therefore$ Translation of $2a$ units in the negative $x$ direction

N2021 P1 Q6

N2021 P1 Q6

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