[GMATH practice question]
Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum possible vertical coordinate. What is the sum of the coordinates of P?
(A) 5
(B) 5.5
(C) 6
(D) 6.5
(E) 7
Answer: [spoiler]__(C)____[/spoiler]
Point P is the point of the circle x^2 + y^2 -2x -4y = 4 wit
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Last edited by fskilnik@GMATH on Fri Sep 14, 2018 8:50 am, edited 1 time in total.
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\[P = \left( {{x_P}\,,\,\,{y_P}} \right)\,\,\, \in \,\,\,\,\left\{ {\,\left( {x,y} \right)\,\,\,:\,\,\,{x^2} - 2x + {y^2} - 4y = 4\,} \right\}\]fskilnik wrote: [GMATH practice question]
Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum possible vertical coordinate. What is the sum of the coordinates of P?
(A) 5
(B) 5.5
(C) 6
(D) 6.5
(E) 7
\[{y_P}\,\,\max \,\,\,,\,\,\,\,\,? = {x_P} + {y_P}\]
Let´s apply the "filling the squares" technique presented in our course!
\[{x^2} - 2x + {y^2} - 4y = 4\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\underbrace {{x^2} - 2x + \underline 1 }_{{{\left( {x - 1} \right)}^{\,2}}} + \underbrace {{y^2} - 4y + \underline 4 }_{{{\left( {y - 2} \right)}^{\,2}}} = \underbrace {4 + \underline 1 + \underline 4 }_9\]
\[P\,\, \in \,\,\,\left\{ {\,\,\left( {x,y} \right)\,\,:\,\,\,{{\left( {x - 1} \right)}^2} + {{\left( {y - 2} \right)}^2} = {3^2}} \right\}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,P\,\, \in \,\,\,\, \odot \,\,\left\{ \begin{gathered}
\,{\text{Centre}}\, = \left( {1,2} \right) \hfill \\
{\text{Radius}} = 3 \hfill \\
\end{gathered} \right.\]
\[\left. \begin{gathered}
P = \left( {{x_P}\,,\,\,{y_P}} \right)\,\, \in \,\,\,\, \odot \,\, \hfill \\
{y_P}\,\,\max \,\, \hfill \\
\end{gathered} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{geometrically}}\,\,{\text{evident}}\,!} \,\,\,\,\,P = \left( {1,2 + 3} \right) = \left( {1,5} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 6\]
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Last edited by fskilnik@GMATH on Sat Sep 15, 2018 4:00 pm, edited 2 times in total.
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Reorganizing the problem: y^2 - 4y +x^2- 2x = 4fskilnik wrote:[GMATH practice question]
Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum possible vertical coordinate. What is the sum of the coordinates of P?
(A) 5
(B) 5.5
(C) 6
(D) 6.5
(E) 7
Answer: [spoiler]__(C)____[/spoiler]
To maximize the vertical is to maximize y. To maximize y, we need to minimize the expression x^2 - 2x since to the extent it is positive will reduce the right side, 4, available for the y expression to satisfy.
Looking at x^2 - 2x it is tempting to believe the minimum value is 0, but can we do better than that ?
If we set x=1, then the expression reduces to -1, that's better than 0. If we set x=2, the expression = 0, so headed in the wrong direction.
If we set x = 0, the expression = 0 , again the wrong direction. Testing x=-1 the expression = 3.
So x = 1 minimizes the expression. Plugging x=1 into the expression and solving for y:
y^2-4y + 1 - 2 = 4 > y^2-4y - 5 =0
Solving for Y: (y+1)(y-5) = 0 so y = -1, 5. The maximum is 5.
So the coordinates of the point P where y is a maximum, 5, are (1,5) so the sum of the coordinates is 6, C
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Alternate solution (adapted from the very nice idea of regor60, posted above. Thank you for your contribution!):fskilnik wrote: [GMATH practice question]
Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum possible vertical coordinate. What is the sum of the coordinates of P?
(A) 5
(B) 5.5
(C) 6
(D) 6.5
(E) 7
\[P = \left( {{x_P}\,,\,\,{y_P}} \right)\,\,\, \in \,\,\,\,\left\{ {\,\left( {x,y} \right)\,\,\,:\,\,\,{x^2} - 2x + {y^2} - 4y = 4\,} \right\}\]
\[{y_P}\,\,\max \,\,\,,\,\,\,\,\,? = {x_P} + {y_P}\]
\[{x^2} - 2x + {y^2} - 4y = 4\,\,\,\,\, \Leftrightarrow \,\,\,\,{y^2} - 4y = 4 - \left( {{x^2} - 2x + \underline 1 } \right) + \underline 1 = 5 - {\left( {x - 1} \right)^2}\]
\[{y^2} - 4y = 5 - {\left( {x - 1} \right)^2} \leqslant 5\]
\[y\,\,\max \,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}
{x = {x_p} = 1} \\
{{y_p}^2 - 4{y_p} = 5}
\end{array}\begin{array}{*{20}{c}}
{} \\
{\,\,\,\mathop \Rightarrow \limits^{S = 4\,,\,P = - 5} \,\,\,\,{y_p} = \max \left\{ {5, - 1} \right\}\,\, = 5\,\,\,\,}
\end{array}} \right.\]
\[? = 1 + 5 = 6\]
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Last edited by fskilnik@GMATH on Fri Sep 14, 2018 12:58 pm, edited 1 time in total.
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(x-h)² + (y-k)² = r² is a circle with a center at (h, k) and a radius of r.fskilnik wrote:[GMATH practice question]
Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum possible vertical coordinate. What is the sum of the coordinates of P?
(A) 5
(B) 5.5
(C) 6
(D) 6.5
(E) 7
x² + y² - 2x - 4y = 4
x² - 2x + y² - 4y = 4
x² - 2x + 1 + y² - 4y + 4 = 4 + 1 + 4
(x-1)² + (y-2)² = 9
(x-1)² + (y-2)² = 3³
The equation above constitutes a circle with a center at (1, 2) and a radius of 3.
Since r=3, the highest point is located exactly 3 units above the center:
As shown in the figure above, the highest point is at (1, 5).
Sum of the coordinates = 1+5 = 6.
The correct answer is C.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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