Course:MATH102/Question Challenge/2008 December Q1.3

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Question

1.3: Consider a point in the first quadrant on the hyperbola

$\displaystyle x^{2}-y^{2}=1$

with $x=2$ . The slope of the tangent line at that point is

(a) $2/{\sqrt {5}}$

(b) $1/{\sqrt {3}}$

(c) ${\sqrt {5}}/2$

(d) $2/{\sqrt {3}}$

(e) $2/3$

Hints

Hint
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Solutions

Solution

$x^{2}-y^{2}=1$ We will use implicit differentiation to find the first derivative: $2x-2y\cdot {\frac {dy}{dx}}=0$ From this we can conclude that ${\frac {dy}{dx}}={\frac {2x}{2y}}={\frac {x}{y}}$ We will solve for $y$ at the point $x=2$: $x^{2}-y^{2}=1$ $(2)^{2}-y^{2}=1$ $4-y^{2}=1$ $y^{2}=4-1$ $y^{2}=3$ $y={\sqrt {3}}$ We can substitute the values $x=2,y={\sqrt {3}}$ into the expression found above for $dy/dx$. Alternatively, we could have first substituted in $x=2$ and $y={\sqrt {3}}$ and only then isolated $dy/dx$ to find the slope of the tangent line: $2x-2y\cdot {\frac {dy}{dx}}=0$ $2(2)-2({\sqrt {3}})\cdot {\frac {dy}{dx}}=0$ $4-2({\sqrt {3}})\cdot {\frac {dy}{dx}}=0$ $-2({\sqrt {3}})\cdot {\frac {dy}{dx}}=-4$ ${\frac {dy}{dx}}={\frac {4}{2({\sqrt {3}}}}$ ${\frac {dy}{dx}}={\frac {2}{\sqrt {3}}}$ The correct answer is D)