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10.2 parabola,ellipse,hyperbola — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHS10.2 parabola,ellipse,hyperbola1 Mark
MCQ
For the hyperbola $\frac{{{x^2}}}{9} - \frac{{{y^2}}}{3} = 1$  the incorrect statement is :
  • A
    product of the perpendicular distances from any point on the hyperbola on its asymptotes is less than the length of its latus rectum .
  • B
    its eccentricity is $4/3$
  • C
    length of the latus rectum is $2$
  • ✓
    both $(A)$ and $(B)$
✓

Answer

Correct option: D.
both $(A)$ and $(B)$
d
$p_1 p_2 = $ $\frac{{{a^2}\,{b^2}}}{{{a^2}\,\, + \,\,{b^2}}}$ ; $e = \sec \theta$ 
$e^2 =1 +\frac{9}{3}\,$ $= 4$  $ \Rightarrow $ $e = 2 = \sec\theta $ ($B$  is correct)
$\Rightarrow$ $\theta = 60°$
angle between the two asymptotes is $120°$ 
$\Rightarrow $ acute angle is $60°$ $ \Rightarrow $ $(A)$  is correct  
$C :$ $LLR =$ $\frac{{2{b^2}}}{a}\,$  = $2.\frac{3}{3}\,\,$ = $2 $ $\Rightarrow$  $(C)$ is correct  
$p_1p_2 $ = $\frac{{ab(\sec \theta  + \tan \theta )}}{{\sqrt {{a^2} + {b^2}} }}\,\,\,\frac{{ab(\sec \theta  - \tan \theta )}}{{\sqrt {{a^2} + {b^2}} }}$
$=$ $\frac{{{a^2}\,{b^2}}}{{{a^2} + {b^2}}}\,({\sec ^2}\theta  - {\tan ^2}\theta )\,\, = \,\,\frac{{9\,.\,3}}{{12}}\,\, = \,\,\,\frac{9}{4}\,$  

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