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A curve is represented by the equation x^2 y^3 = k^3, where

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A curve is represented by the equation x^2 y^3 = k^3, where

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Source: e-GMAT

A curve is represented by the equation x^2 y^3=k^3, where k < 0. At how many points does the line, y = -a, where a is an integer, intersects this curve?

A. 0
B. 1
C. 2
D. 4
E. Cannot be determined

The OA is E.

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BTGmoderatorLU wrote:
Source: e-GMAT

A curve is represented by the equation x^2 y^3=k^3, where k < 0. At how many points does the line, y = -a, where a is an integer, intersects this curve?

A. 0
B. 1
C. 2
D. 4
E. Cannot be determined
The correct answer is (E):

\[\left\{ \begin{gathered}
\,{x^2}{y^3} = {k^3}\,\,\,\left( {k < 0} \right) \hfill \\
y = - a\,\,\,\left( {a\,\,\operatorname{int} } \right) \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,{x^2}{\left( { - a} \right)^3} = {k^3}\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,{x^2} = - {\left( {\frac{k}{a}} \right)^3}\,\,\,\,\,,\,\,\,\,a \ne 0\]

Take (for instance) :

(a,k) = (1,-1) , then we have 2 points (x,y) = (x, -a) of intersection: (-1, -1) and (1, -1)
(a,k) = (-1,-1) , then we have 0 points (x,y) = (x,-a) of intersection, because x^2 = -1 has no solutions (in the real numbers)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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BTGmoderatorLU wrote:
Source: e-GMAT

A curve is represented by the equation x^2 y^3=k^3, where k < 0. At how many points does the line, y = -a, where a is an integer, intersects this curve?

A. 0
B. 1
C. 2
D. 4
E. Cannot be determined

The OA is E.
Let's plug-in y = - a in the equation x^2*y^3 = k^3

-x^2*a^3 = -|k^3|; we are given that k < 0

x^2*a^3 = |k^3|

x^2 = |k^3| / a^3

There are three cases...

1. If a > 0 then x^2 = |k^3| / |a^3| => x = ±√(|k/a|^3. There are two points.

2. If a = 0 then x^2 = |k^3| / |0^3| is indeterminable.

3. If a ≤ 0 then x^2 = |k^3| / -|a^3| => x^2 = -[|k^3| / |a^3|]; squareroot of a negative number is not possible. Indeterminable.

The correct answer: E

Hope this helps!

-Jay
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