The equation r cos theta = 2 a sin^2 theta re | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the polar equation: $r \cos 	heta = 2a \sin^2 	heta$. We know the conversion relations: $x = r \cos 	heta$,$y = r \sin 	heta$,and $r^2 =

Vedclass · Accepted Answer

$x^3 = y^2(2a - x)$

Vedclass · Answer

(C) Given the polar equation: $r \cos 	heta = 2a \sin^2 	heta$. We know the conversion relations: $x = r \cos 	heta$,$y = r \sin 	heta$,and $r^2 = x^2 + y^2$. From the given equation,we have $r \cos 	heta = 2a \sin^2 	heta$. Multiply both sides by $r^2$: $r^3 \cos 	heta = 2a (r \sin 	heta)^2$. Since $r^2 = x^2 + y^2$,we have $r = \sqrt{x^2 + y^2}$. Substituting $x = r \cos 	heta$ and $y = r \sin 	heta$: $r^2 (r \cos 	heta) = 2a (r \sin 	heta)^2$ $(x^2 + y^2)x = 2a y^2$. Rearranging the terms: $x(x^2 + y^2) = 2a y^2$ $x^3 + xy^2 = 2a y^2$ $x^3 = 2a y^2 - xy^2$ $x^3 = y^2(2a - x)$. Thus,the correct option is $C$.

The equation $r \cos \theta = 2 a \sin^2 \theta$ represents the curve

Similar Questions

The number of solid cones with integer radius and integer height each having its volume numerically equal to its total surface area is

Find the locus of a point from which the lengths of the tangents drawn to the two circles $x^2 + y^2 - 5x - 3 = 0$ and $3x^2 + 3y^2 + 2x + 4y - 6 = 0$ are equal.

The locus of the centre of the circle touching the $x$-axis and passing through the point $(-1, 1)$ is

From a point $A(0,3)$ on the circle $(x+2)^2+(y-3)^2=4$,a chord $AB$ is drawn and it is extended to a point $Q$ such that $AQ=2AB$. Then the locus of $Q$ is

Let $PQ$ and $MN$ be two straight lines touching the circle $x^{2}+y^{2}-4x-6y-3=0$ at the points $A$ and $B$ respectively. Let $O$ be the centre of the circle and $\angle AOB=\pi/3$. Then the locus of the point of intersection of the lines $PQ$ and $MN$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module