For hyperbola frac{{{x^2}}}{9} - frac{{{y^2}}}{3} = 1 incorrect statement is :

Gujarati

Hindi

10-2. Parabola, Ellipse, Hyperbola

normal

For the hyperbola $\frac{{{x^2}}}{9} - \frac{{{y^2}}}{3} = 1$ the incorrect statement is :

product of the perpendicular distances from any point on the hyperbola on its asymptotes is less than the length of its latus rectum .

its eccentricity is $4/3$

length of the latus rectum is $2$

both $(A)$ and $(B)$

Solution

$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 $ = $\frac{{ab(\sec \theta + \tan \theta )}}{{\sqrt {{a^2} + {b^2}} }}\,\,\,\frac{{ab(\sec \theta – \tan \theta )}}{{\sqrt {{a^2} + {b^2}} }}$
$=$ $\frac{{{a^2}\,{b^2}}}{{{a^2} + {b^2}}}\,({\sec ^2}\theta – {\tan ^2}\theta )\,\, = \,\,\frac{{9\,.\,3}}{{12}}\,\, = \,\,\,\frac{9}{4}\,$

Standard 11

Mathematics

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $16 x^{2}-9 y^{2}=576$

medium

Let the foci of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{7}=1$ and the hyperbola $\frac{ x ^{2}}{144}-\frac{ y ^{2}}{\alpha}=\frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:-

medium

(JEE MAIN-2022)

The eccentricity of the hyperbola can never be equal to

easy

If the circle $x^2 + y^2 = a^2$ intersects the hyperbola $xy = c^2 $ in four points $ P(x_1, y_1), Q(x_2, y_2), R(x_3, y_3), S(x_4, y_4), $ then