Math, asked by sharmaanjana9900, 8 months ago

Trace the curve y² (a²+ x²) = x²(a²-x²), a>0​

Answers

Answered by ranoutnikhil987
1

Answer:

The best way is to do it with the help of polar coordinates.

{

x

=

r

cos

θ

y

=

r

sin

θ

substituting

r

2

r

cos

θ

=

a

r

2

sin

2

θ

or for

r

>

0

r

=

a

sin

θ

tan

θ

we can compute easily some values for

r

given

θ

(

θ

0

π

4

π

2

r

0

2

infinity

)

We know also that the curve is symmetric regarding the

x

=

0

axis having also a vertical asymptote at

x

=

a

because

y

2

=

x

3

a

x

Attached a plot for

a

=

2

Answered by ravilaccs
1

Answer:

The tangents to the curve at (a, 0) and (-a,0) are parallel to Y-axis.

The curve has no asymptotes.

Step-by-step explanation:

  • A straight line is said to be an asymptote to an infinite branch of a curve if the perpendicular distance from a point on the curve to the given line approaches to zero as the point moves to infinity along the branch of curve.
  • The asymptotes parallel to x-axis are called horizontal asymptotes, those which are parallel to y-axis are called  vertical asymptotes and those which are neither parallel to x-axis nor parallel to y-axis are called oblique asymptotes.
  • The equations of a vertical asymptotes are obtained by equating the coefficient of highest degree term in y to zero if it is not a constant.
  • To obtain the equations of oblique asymptotes, substitutes y = mx + c in the given equation. then equate the coefficients of the highest degree term in x and next highest degree term in x to zero, if it is not a constant, to determine m and c. If the values of m and c exists, then y = mx + c is the equation of the oblique asymptote.

(i) Symmetry: Symmetrical about both the axes.

(ii) Points: Passes through (0, 0), (a, 0) and (-a,0). Loop between (0, 0) and (a, 0); also between (0, 0) and (-a, 0).

(iii) Tangents: Lowest degree terms a2y2-x2a2 = 0, we get y = ± x, the tangents.

(iv) Region:

y= +x\sqrt{\frac{a^2-x^2}{a^2+x^2} } \\y= -x\sqrt{\frac{a^2-x^2}{a^2+x^2} }

Note that y is imaginary if |x|>a. Hence the whole curve lies between x = a and x = -a.

Therefore the tangents to the curve at (a, 0) and (-a,0) are parallel to Y-axis.

The curve has no asymptotes.

Reference Link

  • https://brainly.in/question/12435862
  • https://brainly.in/question/314520
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