Line segments AK1,AK2,.....,AKn are drawn from A(1,1) where K1,K2,K3,.....,Kn are points in 1st quadrant on (x−1)2a2+(y−1)2b2=1(a>b) such that chord AKr makes an angle of rπ2n with the positive x-axis. The value of limn→∞(1n(∑r=1n(AKr)2)) is

(x+1)2a2+(y−1)2b2=1The co-ordinates of the point Kr will be (1+Rcosθ,1+Rsinθ), where R is AKr and θ=rπ2n⇒R2cos2θa2+R2sin2θb2=1 limb→∞(1n(∑x−1n(AKr)2)) ⇒limb→∞1n∑x−1n1cos2(rπ/2n)a2+sin2(rπ/2n)b2=∫01a2b2b2cos2(π2x)+a2sin2(π2x)dx

Line segments AK1,AK2,.....,AKn are drawn from A(1,1) where K1,K2,K3,.....,Kn are points in 1st quadrant on (x−1)2a2+(y−1)2b2=1(a>b) such that chord AKr makes an angle of rπ2n with the positive x-axis. The value of limn→∞(1n(∑r=1n(AKr)2)) is

(x+1)2a2+(y−1)2b2=1The co-ordinates of the point Kr will be (1+Rcosθ,1+Rsinθ), where R is AKr and θ=rπ2n⇒R2cos2θa2+R2sin2θb2=1 limb→∞(1n(∑x−1n(AKr)2)) ⇒limb→∞1n∑x−1n1cos2(rπ/2n)a2+sin2(rπ/2n)b2=∫01a2b2b2cos2(π2x)+a2sin2(π2x)dx

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Line segments $A K_{1}, A K_{2}, ....., A K_{n}$ are drawn from $A (1, 1)$ where $K_{1}, K_{2}, K_{3}, ....., K_{n}$ are points in $1^{s t}$ quadrant on $\frac{{(x - 1)}^{2}}{a^{2}} + \frac{{(y - 1)}^{2}}{b^{2}} = 1 (a > b)$ such that chord $A K_{r}$ makes an angle of $\frac{r π}{2 n}$ with the positive x-axis. The value of $\lim_{n \to \infty} (\frac{1}{n} (\sum_{r = 1}^{n} {(A K_{r})}^{2}))$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{1}{a b}$

$\frac{π}{2} a b$

$\frac{2}{π} a b$

$a b$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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

The co-ordinates of the point $K_{r}$ will be $(1 + R \cos θ, 1 + R \sin θ),$ where R is $A K_{r}$ and $θ = \frac{r π}{2 n} \Rightarrow \frac{R^{2} \cos^{2} θ}{a^{2}} + \frac{R^{2} \sin^{2} θ}{b^{2}} = 1$
$\lim_{b \to \infty} (\frac{1}{n} (\sum_{x - 1}^{n} {(A K_{r})}^{2}))$
$\Rightarrow \lim_{b \to \infty} \frac{1}{n} \sum_{x - 1}^{n} \frac{1}{\frac{\cos^{2} (r π / 2 n)}{a^{2}} + \frac{\sin^{2} (r π / 2 n)}{b^{2}}} = \int_{0}^{1} \frac{a^{2} b^{2}}{b^{2} \cos^{2} (\frac{π}{2} x) + a^{2} \sin^{2} (\frac{π}{2} x)} d x$