JEE MATHEMATICS MCQ: ELLIPSE

1 Hard

1 of 1

Analyzing +4 | -1

The curve represented by the equation

\frac{x^2}{\sin \sqrt{2}-\cos \sqrt{3}}+\frac{y^2}{\sin \sqrt{3}-\cos \sqrt{2}}=1

View answer and explanation

Correct option: A

An ellipse with foci on

X

-axis

Option A (An ellipse with foci on

X

-axis) is correct because both denominators are positive and the denominator of

x^2

is greater than the denominator of

y^2

More Questions from ELLIPSE

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