Find the equations for (a) the tangent lines, and (b) the normal lines, to the hyperbola $y^2/4 - x^2/2 = 1$ when $x = 4$.
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i used the x^2/a^a- y^2/b^2, but its kinda conflicting the initial equation given. – precious May 14 '13 at 10:57
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If $x=4, \frac{y^2}4=1+\frac{4^2}2=9\implies y=\pm6$
Using Article 305 of this, the tangent of $$\frac{y^2}4-\frac{x^2}2=1$$ at $(h,k)$ is $$ \frac{y\cdot k}4-\frac{x\cdot h}2=1$$
Do you know how to find the perpendicular of a given line from a given point $(4,\pm 6)$?
lab bhattacharjee
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is the equation for the tangent line will be same as above where i substitute the values for x n y ryt? – precious May 14 '13 at 11:00
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help me understand please. i need to write out the equation for the tangent line n the normal line – precious May 14 '13 at 11:03
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@precious, we need replace $(h,k)$ with $(4,6)$ and $(4,-6)$ one by one – lab bhattacharjee May 14 '13 at 11:04
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thats gonna give me a positive n negative equation. is this equation 4 the tangent then? – precious May 14 '13 at 11:06
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$$\frac{y\cdot 6}4-\frac{x\cdot 4}2=1$$ and $$\frac{y\cdot (-6)}4-\frac{x\cdot 4}2=1$$. Please simplify – lab bhattacharjee May 14 '13 at 11:13
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@precious, the gradient of $$\frac{y\cdot 6}4-\frac{x\cdot 4}2=1\iff y=\frac43x+\frac23 $$ is $\frac43$. So, the gradient of the normal is $\frac{-1}{\frac43}=-\frac34$
So, the equation of the normal which passes through $(4,6)$ will be $$\frac{y-6}{x-4}=-\frac34$$
– lab bhattacharjee May 16 '13 at 08:01