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

If the eccentricity and the length of the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are $\frac{\sqrt{3}}{2}$ and $1$ respectively,then the sum of the lengths of the major axis and minor axis of the ellipse is

In an ellipse,two vertices are $(5,0)$ and $(0,-4)$. Then the equation of the ellipse is

The locus of the midpoints of the intercepted portion of the tangents by the coordinate axes,which are drawn to the ellipse $x^2+2y^2=2$,is

The eccentric angle of the point $P(-6, 2)$ on the ellipse $\frac{x^2}{48} + \frac{y^2}{16} = 1$ is: (in $^{\circ}$)

The locus of the point of intersection of the perpendicular tangents to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module