A normal is drawn to the ellipse x^29+y^2=1 at the point 3cos⁡θ,sin⁡θ where 0

Question

A normal is drawn to the ellipse x^29+y^2=1 at the point 3cos⁡θ,sin⁡θ where 0<θ<(π)/2. If N is the foot of the perpendicular from the origin O to the normal such that ON=2, then θ is equal to

MARKS Content Team · Accepted Answer

Correct answer: (D) (π)/6. Equation of the normal at 3cosθ,sinθ is 3x/(cos⁡θ)-y/(sin⁡θ)=8 So, ON=(-8√(9sec^2⁡θ+cosec^2⁡θ)=2 (given)⇒16=91+tan^2⁡θ+1+cot^2⁡θ⇒6=6+3tan⁡θ-cot⁡θ^2⇒3tan⁡θ-cot⁡θ=0Hence, tan^2⁡θ=1)/3⇒θ=(π)/6

A normal is drawn to the ellipse $\frac{x^{2}}{9} + y^{2} = 1$ at the point $(3 \cos θ, \sin θ)$ where $0 < θ < \frac{π}{2} .$ If $N$ is the foot of the perpendicular from the origin $O$ to the normal such that $O N = 2$ , then $θ$ is equal to

Solution

More Ellipse Questions

A normal is drawn to the ellipse x29+y2=1 at the point 3cos⁡θ,sin⁡θ where 0<θ<π2. If N is the foot of the perpendicular from the origin O to the normal such that ON=2, then θ is equal to

Solution

More Ellipse Questions

A normal is drawn to the ellipse $\frac{x^{2}}{9} + y^{2} = 1$ at the point $(3 \cos θ, \sin θ)$ where $0 < θ < \frac{π}{2} .$ If $N$ is the foot of the perpendicular from the origin $O$ to the normal such that $O N = 2$ , then $θ$ is equal to