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09 Aug 2020, 17:23
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\(\triangle ABC\) has coordinates A= (-4, 0), B= (4 , 0), and C= (0 , 3). Let P be the point in the first quadrant such that \(\triangle ABP\) has half the area of \(\triangle ABC\) but both triangles have the same perimeter. What is the length of CP? 
hyrCFIlaO6-92280.png [ 8.29 KiB | Viewed 7747 times ]
A. \(\sqrt{15}\)
B. 4
C. \(\sqrt{17}\)
D. \(\sqrt{21}\)
E. 5
Attachment:
hyrCFIlaO6-92280.png [ 8.29 KiB | Viewed 7747 times ]
A. \(\sqrt{15}\)
B. 4
C. \(\sqrt{17}\)
D. \(\sqrt{21}\)
E. 5
10 Aug 2020, 15:33
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Lipun
1 number bhai! But we don't need to know ellipse to answer this question.
Suppose coordinates of P =(x,y)
Since the △ABP has half the area of △ABC and both triangle have common base, y must be equal to 3/2.
Also, AC=BC=5, hence AP+PB = 5+5=10........(1) [perimeter of both triangles are equal]
\(AP^2 = (4+x)^2+(\frac{3}{2})^2\).......(2)
\(PB^2= (x-4)^2+(\frac{3}{2})^2\)......(3)
\((10-AP)^2=(x-4)^2+(\frac{3}{2})^2\) ..........(4)
Subtract (2) and (4)
\(AP = 5+ \frac{4x}{5}\)
Put value of AP in equation 2
\(x^2 = \frac{75}{4}\)
\(CP^2 = \frac{75}{4}+\frac{9}{4} = 21\)
\(CP = \sqrt{21}\)
1 number bhai! But we don't need to know ellipse to answer this question.
Suppose coordinates of P =(x,y)
Since the △ABP has half the area of △ABC and both triangle have common base, y must be equal to 3/2.
Also, AC=BC=5, hence AP+PB = 5+5=10........(1) [perimeter of both triangles are equal]
\(AP^2 = (4+x)^2+(\frac{3}{2})^2\).......(2)
\(PB^2= (x-4)^2+(\frac{3}{2})^2\)......(3)
\((10-AP)^2=(x-4)^2+(\frac{3}{2})^2\) ..........(4)
Subtract (2) and (4)
\(AP = 5+ \frac{4x}{5}\)
Put value of AP in equation 2
\(x^2 = \frac{75}{4}\)
\(CP^2 = \frac{75}{4}+\frac{9}{4} = 21\)
\(CP = \sqrt{21}\)
General Discussion
10 Aug 2020, 07:21
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Points C and P lie on the circumference of an ellipse.
The equation of the ellipse is given by \(\frac{x^2}{5^2} + \frac{y^2}{3^2} = 1\)
Substitute \(y=\frac{3}{2}\) in the equation. This will yield \(x^2 = \frac{75}{4}\)
Now, \(CP^2 = x^2 + y^2 = \frac{75}{4} + \frac{9}{4} = 21\)
=> \(CP = \sqrt{21}\)
nick1816, do we need to know the equation of an ellipse to answer this? Is there any other approach?
Thanks
Lipun
The equation of the ellipse is given by \(\frac{x^2}{5^2} + \frac{y^2}{3^2} = 1\)
Substitute \(y=\frac{3}{2}\) in the equation. This will yield \(x^2 = \frac{75}{4}\)
Now, \(CP^2 = x^2 + y^2 = \frac{75}{4} + \frac{9}{4} = 21\)
=> \(CP = \sqrt{21}\)
nick1816, do we need to know the equation of an ellipse to answer this? Is there any other approach?
Thanks
Lipun
