For the hyperbola x^2 9 - y^2 3 = 1 the incorrect statement is : - Vidyadip

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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