The curve represented by x = 5(cos t + sin t) | vedclass

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

Question

(C) Given equations are $x = 5(\cos t + \sin t)$ and $y = 3(\cos t - \sin t)$.Dividing by the coefficients,we get $\frac{x}{5} = \cos t + \sin t$ and

Vedclass · Accepted Answer

ellipse

Vedclass · Answer

(C) Given equations are $x = 5(\cos t + \sin t)$ and $y = 3(\cos t - \sin t)$.Dividing by the coefficients,we get $\frac{x}{5} = \cos t + \sin t$ and $\frac{y}{3} = \cos t - \sin t$.Squaring both equations:$(\frac{x}{5})^2 = (\cos t + \sin t)^2 = \cos^2 t + \sin^2 t + 2 \sin t \cos t = 1 + \sin(2t)$.$(\frac{y}{3})^2 = (\cos t - \sin t)^2 = \cos^2 t + \sin^2 t - 2 \sin t \cos t = 1 - \sin(2t)$.Adding these two results:$(\frac{x}{5})^2 + (\frac{y}{3})^2 = (1 + \sin(2t)) + (1 - \sin(2t)) = 2$.Thus,$\frac{x^2}{25} + \frac{y^2}{9} = 2$,which represents an ellipse.

The curve represented by $x = 5(\cos t + \sin t)$ and $y = 3(\cos t - \sin t)$ is (where $t$ is a parameter):

Similar Questions

If a tangent having slope $m = \frac{1}{3}$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ is a normal to the circle $(x + 1)^2 + (y + 1)^2 = 1$,then $a^2$ lies in the interval:

The locus of the foot of the perpendicular drawn from the centre of the ellipse $x^2 + 3y^2 = 6$ to any tangent to it is:

If tangents are drawn to the ellipse $x^2+2y^2=2$,then the locus of the midpoints of the intercepts made by the tangents between the coordinate axes is

The eccentricity of the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$ is

The eccentricity of the curve represented by the equation ${x^2} + 2{y^2} - 2x + 3y + 2 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module