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kez.b · Dec 5, 2010, 11:59 PM

Unknown008 · Dec 6, 2010, 02:44 AM

I would have done it this way:

Find the derivative of the curve.

Find the gradient of the curve as x tends to infinity.
Since the curve is symmetrical, you'll have a second gradient, which is the negative of the first gradient.

Find the normal to these gradients.

And find the points on the curve where the gradient is equal to the gradient of the normal.

You then have a point and a gradient for each line, which enables you to get the equation of the tangent.

galactus · Dec 6, 2010, 03:02 PM

If we have the hyperbola:

$\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$

the vertices of the hyperbola lie on the y-axis and the asymptotes of the hyperbola are the lines with equation

$y=\pm\frac{a}{b}x$

Your hyperbola can be written as:

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

Remember, perpendicular lines have slopes which are negative reciprocals of the asymptotes' slopes.

Here is the graph. The red lines indicate the asymptotes.