<p>The curve $C$ has equation $y=\sqrt{5{{x}^{2}}+4}$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Sketch curve $C$, indicating clearly the axial intercepts, the equations of the asymptotes and the coordinates of the stationary points.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence, by inserting a suitable graph, determine the range of values of $h$, where $h$ is a positive constant, such that the equation $\sqrt{5{{x}^{2}}+4}=h\sqrt{(1-{{x}^{2}})}$ has no real roots.</span></div></li></ol>

The curve $C$ has equation $y=\sqrt{5{{x}^{2}}+4}$.

Sketch curve $C$, indicating clearly the axial intercepts, the equations of the asymptotes and the coordinates of the stationary points.

Hence, by inserting a suitable graph, determine the range of values of $h$, where $h$ is a positive constant, such that the equation $\sqrt{5{{x}^{2}}+4}=h\sqrt{(1-{{x}^{2}})}$ has no real roots.

Parametric Equations

Graphing Techniques

International Baccalaureate Higher Level Mathematics

IB HL Math

2013 MJC PROMO Q6

$y^2 = 5x^2 + 4$

$y^2 - 5x^2 = 4$

$\frac{y^2}{4} - \frac{5x^2}{4} = 1$

$\frac{y^2}{2^2} - \frac{x^2}{(\frac{2}{\sqrt{5}})^2} = 1$

**Asymptotes:** $\frac{y^2}{4} = \frac{x^2}{\frac{4}{5}}$

$y^2 = 5x^2$

$y = \sqrt{5}x \quad \text{or} \quad y = -\sqrt{5}x$

Graph shows hyperbola with vertices at $(0, 2)$ and $(0, -2)$, opening upward/downward with asymptotes $y = \pm\sqrt{5}x$.

$y = h\sqrt{1 - x^2}$

$\left(\frac{y}{h}\right)^2 = 1 - x^2$

$\frac{y^2}{h^2} + x^2 = 1$

$\frac{y^2}{h^2} + \frac{x^2}{1^2} = 1$

This is an ellipse with centre $(0, 0)$, semi-major axis $h$ (vertical), semi-minor axis $1$ (horizontal).

$\therefore$ For $y = h\sqrt{1-x^2}$ to cut/intersect $C$:

$0 < h < 2$

2013 MJC PROMO Q6

2013 MJC PROMO Q6

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