Question

Find the vertices,eccentricity,foc and asymptotes of the hyperbola x^2/8-y^2/4=1.Also trace it.Under what condition on ⋋ the line x+⋋y=2 will be tangent to this hyperbola?explain geometrically

Answer 1

Answer to Question #87969 – Math – Analytic Geometry

Question

Find the vertices, eccentricity, foc and asymptotes of the hyperbola $x^2/8-y^2/4=1$. Also trace it. Under what condition on $\succ$ the line $x+\succ y=2$ will be tangent to this hyperbola? Explain geometrically

Solution

$$\frac{x^2}{8} - \frac{y^2}{4} = 1$$

Rewrite in the form of $a$ standard hyperbola equation

$$\frac{(x - 0)^2}{(2\sqrt{2})^2} - \frac{(y - 0)^2}{2^2} = 1$$ $$\therefore (h, k) = (0, 0), \, a = 2\sqrt{2}, \, b = 2$$ $$\text{Vertices} = (0, 0)$$

The eccentricity $= e \frac{\sqrt{a^2 + b^2}}{a}$

$$e = \frac{\sqrt{(2\sqrt{2})^2 + 2^2}}{2\sqrt{2}} = \sqrt{\frac{3}{2}}$$

For right-left hyperbola the asymptotes are $y = \pm \frac{b}{a} (x - h) + k$

$$y = \frac{2}{2\sqrt{2}} (x - 0) + 0, \quad y = -\frac{2}{2\sqrt{2}} (x - 0) + 0$$ $$y = \frac{x}{\sqrt{2}}, \quad y = -\frac{x}{\sqrt{2}}$$

The Foci are $(h + c, k), (h - c, k)$, where $c = \sqrt{a^2 + b^2}$

$$c = \sqrt{(2\sqrt{2})^2 + 2^2} = 2\sqrt{3}$$

Then, foci $= (0 + 2\sqrt{3}, 0), (0 - 2\sqrt{3}, 0)$

$$= (2\sqrt{3}, 0), (-2\sqrt{3}, 0)$$

Slope of tangent to curve $= \frac{dy}{dx}$

On differentiating,

$$\frac{x^2}{8} - \frac{y^2}{4} = 1 \dots (i)$$ $$\frac{2x}{8} - \frac{2y}{4} \frac{dy}{dx} = 0$$ $$\frac{dy}{dx} = -\frac{\frac{2x}{8}}{-\frac{2y}{4}} = \frac{x}{2y}$$

From (i), $y = \pm \sqrt{4\left(\frac{x^2}{8} - 1\right)}$

$$y = \pm \sqrt{\frac{x^2}{2} - 4}$$

so, $\frac{dy}{dx} = \frac{x}{\pm 2\sqrt{\frac{x^2}{2} - 4}}$

Given line: $x + \lambda y = 2$

$$y = \frac{-x}{\lambda} + \frac{2}{\lambda}$$

Slope $= \frac{-1}{\lambda}$

We get,

$$\frac{x}{\pm 2\sqrt{\frac{x^2}{2} - 4}} = \frac{-1}{\lambda}$$

Thus, $\lambda = \mp \frac{2\sqrt{\frac{x^2}{2} - 4}}{x}$

Applying these elements and tracing the hyperbola, we get following graph:

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