$x = 4(1 + \cos {\rm{\theta }})$ and $y = 3(1 + \sin {\rm{\theta }})$ are the parametric equations of

$\frac{{{{\left( {x - 3} \right)}^2}}}{9} + \frac{{{{\left( {y - 4} \right)}^2}}}{{16}} = 1$

JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Application +4 | -1

x = 4(1 + \cos \mathrm{\theta })

and

y = 3(1 + \sin \mathrm{\theta })

are the parametric equations of

View answer and explanation

Correct option: D

\frac{{{{\left( {x - 4} \right)}^2}}}{{16}} + \frac{{{{\left( {y - 3} \right)}^2}}}{9} = 1

Option D (

\frac{{{{\left( {x - 4} \right)}^2}}}{{16}} + \frac{{{{\left( {y - 3} \right)}^2}}}{9} = 1

) is correct because it is the equation of the ellipse derived from the given parametric equations.

Practice All Chapter MCQs →

\frac{x^2}{144}+\frac{y^2}{81}=1

m

\mathrm{M}

\mathrm{mM}

P

\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1

S

S'.

A

PSS',

A

A

B

(\sqrt{7}, 0)

(-\sqrt{7}, 0)

P

9 x^{2}+16 y^{2}=144

\mathrm{PA}+\mathrm{PB}

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

r

\frac{x^2}{25}+\frac{y^2}{16}=1

P

x=0

Q

R

Q

y=x

Q

R

N P

Q

P

O