Conic Sections Question 371
Question: If $ P\equiv (x,\ y) $ , $ F_1\equiv (3,\ 0) $ , $ F_2\equiv (-3,\ 0) $ and $ 16x^{2}+25y^{2}=400 $ , then $ PF_1+PF_2 $ equals
[IIT 1998]
Options:
A) 8
B) 6
C) 10
D) 12
Show Answer
Answer:
Correct Answer: C
Solution:
Equation of the curve is $ \frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1 $
therefore $ -5\le x\le 5,-4\le y\le 4 $
$ PF_1+PF_2=\sqrt{[{{(x-3)}^{2}}+y^{2}]}+\sqrt{[{{(x+3)}^{2}}+y^{2}]} $
$ =\sqrt{{{(x-3)}^{2}}+\frac{400-16x^{2}}{25}}+\sqrt{{{(x+3)}^{2}}+\frac{400-16x^{2}}{25}} $
$ =\frac{1}{5}{ \sqrt{(9x^{2}+625-150x)}+\sqrt{(9x^{2}+625+150x)} } $
$ =\frac{1}{5}{ \sqrt{{{(3x-25)}^{2}}}+\sqrt{{{(3x+25)}^{2}}} }=\frac{1}{5}{ 25-3x+3x+25 } $
$ =10 $ , $ (\because 25-3x>0,25+3x>0 $ )
BETA
