Question

Tracing of the curve of the asteroid xy2 = 4a2 (2a-x) is
क्षुद्रग्रह के वक्र का पता रेखणax = y/2a-y) है
Select one:
a. Asymptotes of the curve is y-axis
वक्र के Asymptotesy-axis है
Ob. All the above
उपरोक्त सभी
c. Curve does not pass through origin
वक्र मूल के माध्यम से पारित नहीं करता है
d. Curve is symmetric about x- axis
वक्र दोनोंx-अक्ष के बारे में सममित है​

Answer 1

given,

$xy^2 = 4a^2 (2a-x)$

1) Symmetry about x- axis .

2) Cuts the x-axis of x=2a .

3)

$y^2 =\frac{8a^3 -4a^2x}{x} \\2y \frac{dy}{dx} = \frac{-8a^3}{x} - 4a^2 \\2y dy/dx =-8a^3 / x^2$

shown in fig

Required Area

$=2\int\limits^a _0 {y} \, dx$

limits (2a to 0)

$=2\int\limits^a_0 {2a.\sqrt{\frac{2a-x}{x} } \fr} \, dx \\=4a \int\limits^\frac{pi}{2} _0 {\sqrt{\frac{2a-2asin^2 thita}{2asin^2thita} } } \,*4a(sinthia .costhita) dx \\=16a^2 \int\limits^\frac{pi}{2} _0 {\frac{cos (thita) }{sin(thita)} } \,.sin (thita).cos(thita) dx \\=16a^2 \int\limits^\frac{pi}{2} _0 {cos^2 (thita)} \, dx \\=\frac{16 a^2}{2}\(((thita)+ \frac{sin2(thita)}{2})^\frac{pi}{2}\\=8a^2 [\frac{pi}{2} -0]\\=4(pi) a^2$

hence , the correct option is (d)Curve is symmetric about x- axis.