Consider ellipses E_{k}: kx^{2} + k^{2}y^{2} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The equation of the ellipse is $kx^{2} + k^{2}y^{2} = 1$,which can be written as $\frac{x^{2}}{1/k} + \frac{y^{2}}{1/k^{2}} = 1$.The end points on

Vedclass · Accepted Answer

$3080$

Vedclass · Answer

(A) The equation of the ellipse is $kx^{2} + k^{2}y^{2} = 1$,which can be written as $\frac{x^{2}}{1/k} + \frac{y^{2}}{1/k^{2}} = 1$.The end points on the axes are $(\pm 1/\sqrt{k}, 0)$ and $(0, \pm 1/k)$.The equation of the chord in the first quadrant joining $(1/\sqrt{k}, 0)$ and $(0, 1/k)$ is $\frac{x}{1/\sqrt{k}} + \frac{y}{1/k} = 1$,which simplifies to $\sqrt{k}x + ky = 1$.The radius $r_{k}$ of the circle $C_{k}$ is the perpendicular distance from the origin $(0,0)$ to this line:$r_{k} = \frac{|0 + 0 - 1|}{\sqrt{(\sqrt{k})^{2} + k^{2}}} = \frac{1}{\sqrt{k + k^{2}}}$.Therefore,$\frac{1}{r_{k}^{2}} = k + k^{2}$.We need to calculate $\sum_{k=1}^{20} \frac{1}{r_{k}^{2}} = \sum_{k=1}^{20} (k + k^{2}) = \sum_{k=1}^{20} k + \sum_{k=1}^{20} k^{2}$.Using the summation formulas $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ and $\sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}$ for $n=20$:$\sum_{k=1}^{20} k = \frac{20 	imes 21}{2} = 210$.$\sum_{k=1}^{20} k^{2} = \frac{20 	imes 21 	imes 41}{6} = 10 	imes 7 	imes 41 = 2870$.Thus,$\sum_{k=1}^{20} \frac{1}{r_{k}^{2}} = 210 + 2870 = 3080$.

Similar Questions

The line $y = 2t^2$ intersects the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ in real points if

The equation of the ellipse whose one vertex is $(0, 7)$ and the directrix is $y = 12$ is:

An angle of intersection of the curves,$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and $x^{2}+y^{2}=ab$,where $a > b$,is :

The foci of the ellipse $25x^2 + 4y^2 + 100x - 4y + 100 = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module