Let p be the number of all triangles that can | vedclass

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

Question

(D) The number of triangles that can be formed is $p = {}^{n}C_{3}$.The number of quadrilaterals that can be formed is $q = {}^{n}C_{4}$.Given $p+q =

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(D) The number of triangles that can be formed is $p = {}^{n}C_{3}$.The number of quadrilaterals that can be formed is $q = {}^{n}C_{4}$.Given $p+q = 126$,we have ${}^{n}C_{3} + {}^{n}C_{4} = 126$.Using the identity ${}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r}$,we get ${}^{n+1}C_{4} = 126$.Since ${}^{9}C_{4} = \frac{9 	imes 8 	imes 7 	imes 6}{4 	imes 3 	imes 2 	imes 1} = 126$,we have $n+1 = 9$,which implies $n = 8$.The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{8} = 1$.Here,$a^2 = 16$ and $b^2 = 8$. Since $a^2 > b^2$,the eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$.$e = \sqrt{1 - \frac{8}{16}} = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

Similar Questions

If the perpendicular distance from the focus of an ellipse $\frac{x^2}{9} + \frac{y^2}{b^2} = 1$ $(b < 3)$ to its corresponding directrix is $\frac{4}{\sqrt{5}}$,then the slope of the tangent to this ellipse drawn at $\left(\frac{3}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$ is

The length of the latus rectum of an ellipse is $\frac{1}{3}$ of its major axis. Its eccentricity is:

If $3x + 4y = 12\sqrt{2}$ is a tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{9} = 1$ for some $a \in R$,then the distance between the foci of the ellipse is:

The number of values of $c$ such that the line $y=4x+c$ touches the curve $\frac{x^{2}}{4}+y^{2}=1$ is

Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the length of the latus rectum of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module