If $$x\cos heta- y\sin heta $$= $$\sqrt {{x^2} + {y^2}} $$and $$\frac{{{{\cos }^2} heta }}{{{a^2}}}$$+ $$\frac{{{{\sin }^2} heta }}{{{b^2}}}$$= $$\frac{1}{{{x^2} + {y^2}}}{ ext{,}}$$then the correct relation is?

Question

If $$x\cos 	heta- y\sin 	heta $$= $$\sqrt {{x^2} + {y^2}} $$and $$\frac{{{{\cos }^2}	heta }}{{{a^2}}}$$+ $$\frac{{{{\sin }^2}	heta }}{{{b^2}}}$$= $$\frac{1}{{{x^2} + {y^2}}}{	ext{,}}$$then the correct relation is?

TuteeHUB · Accepted Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

1.	If $$x\cos \theta- y\sin \theta $$= $$\sqrt {{x^2} + {y^2}} $$and $$\frac{{{{\cos }^2}\theta }}{{{a^2}}}$$+ $$\frac{{{{\sin }^2}\theta }}{{{b^2}}}$$= $$\frac{1}{{{x^2} + {y^2}}}{\text{,}}$$then the correct relation is?
A.	$\frac{{{x^2}}}{{{b^2}}} - \frac{{{y^2}}}{{{a^2}}} = 1$$
B.	$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$
C.	$\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1$$
D.	$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$
Answer» C. $\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1$$

If $$x\cos \theta- y\sin \theta $$= $$\sqrt {{x^2} + {y^2}} $$and $$\frac{{{{\cos }^2}\theta }}{{{a^2}}}$$+ $$\frac{{{{\sin }^2}\theta }}{{{b^2}}}$$= $$\frac{1}{{{x^2} + {y^2}}}{\text{,}}$$then the correct relation is?

Discussion

No Comment Found

Related MCQs

Reply to Comment