2 answers:
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Use the chain rule.
Let u = 25sin²(x), such that dy/dx = dy/du · du/dx
Let u = 25sin²(x), such that dy/dx = dy/du · du/dx
Answer:
The graph in the attached figure
Step-by-step explanation:
we have
-----> equation A
-----> equation B
we know that
The solution of the system of equations is the intersection point both graphs
In this problem
The intersection point is
therefore
The solution is the point
using a graphing tool
see the attached figure
We know that
<span>Since the focus and vertex are above and below each other, rather than side by side, I know that this ellipse must be taller than it is wide.
</span>Then
a²<span> will go with the </span>y<span> part of the equation
</span>Also, since the focus is 8 <span>units below the center, then </span><span>c = 8
</span>since the vertex is 17<span> units above, then </span><span>a = 17
</span>The equation b²<span> = a</span>²<span> – c</span>²<span> gives me
</span>b²=17²-8²-----> b²=225
the equation is
y²/a²+x²/b²=1------> y²/289+x²/225=1
the answer is
y²/289+x²/225=1
see the attached figure
<span>Since the focus and vertex are above and below each other, rather than side by side, I know that this ellipse must be taller than it is wide.
</span>Then
a²<span> will go with the </span>y<span> part of the equation
</span>Also, since the focus is 8 <span>units below the center, then </span><span>c = 8
</span>since the vertex is 17<span> units above, then </span><span>a = 17
</span>The equation b²<span> = a</span>²<span> – c</span>²<span> gives me
</span>b²=17²-8²-----> b²=225
the equation is
y²/a²+x²/b²=1------> y²/289+x²/225=1
the answer is
y²/289+x²/225=1
see the attached figure