Conic Sections Question 54
Question: The equation of the common tangent of the parabolas $ x^{2}=108y $ and $ y^{2}=32x $ , is
Options:
A) $ 2x+3y=36 $
B) $ 2x+3y+36=0 $
C) $ 3x+2y=36 $
D) $ 3x+2y+36=0 $
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Answer:
Correct Answer: B
Solution:
$ S_1\equiv x^{2}-108y=0 $
$ T\equiv xx_1-2a(y+y_1)=0 $
therefore $ xx_1-54( y+\frac{x_1^{2}}{108} )=0 $
$ S_2\equiv y^{2}-32x=0 $
$ T\equiv yy_2-2a(x+x_2)=0 $
therefore $ yy_2-16( x+\frac{y_2^{2}}{32} )=0 $
$ \frac{x_1}{16}=\frac{54}{y_2}=\frac{-x_1^{2}}{y_2^{2}}=r $
therefore $ x_1=16r $ and $ y_2=\frac{54}{r} $
$ \frac{-{{(16r)}^{2}}}{{{(54/r)}^{2}}}=r $
therefore $ r=-\frac{9}{4} $
$ x_1=-36,\ y_2=-24,\ y_1=\frac{{{(36)}^{2}}}{108}=12,\ x_2=18 $ . Equation of common tangent $ (y-12)=\frac{-36}{54}(x+36)\Rightarrow 2x+3y+36=0 $
Aliter: Using direct formula of common tangent $ y{b^{1/3}}+x{a^{1/3}}+{{(ab)}^{2/3}}=0 $ , where $ a=8 $ and $ b=27 $ .
Hence the required tangent is $ 3y+2x+36=0 $ .
BETA
