Let E be the ellipse frac{x^2}{16}+frac{y^2}{ | vedclass

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

Question

(C) Let $A = M(P, Q)$ and $B = M(P, Q^{\prime})$.By the midpoint formula,$A = \frac{P+Q}{2}$ and $B = \frac{P+Q^{\prime}}{2}$.The distance between $A$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Let $A = M(P, Q)$ and $B = M(P, Q^{\prime})$.By the midpoint formula,$A = \frac{P+Q}{2}$ and $B = \frac{P+Q^{\prime}}{2}$.The distance between $A$ and $B$ is given by $|A - B| = |\frac{P+Q}{2} - \frac{P+Q^{\prime}}{2}| = |\frac{Q - Q^{\prime}}{2}| = \frac{1}{2} |Q - Q^{\prime}|$.Here,$|Q - Q^{\prime}|$ represents the distance between two points $Q$ and $Q^{\prime}$ on the ellipse $E$.The maximum distance between any two points on the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is the length of the major axis.The major axis length is $2a = 2 	imes 4 = 8$.Therefore,the maximum value of $|Q - Q^{\prime}|$ is $8$.Thus,the maximum distance between $M(P, Q)$ and $M(P, Q^{\prime})$ is $\frac{8}{2} = 4$.

Similar Questions

Find the number of points on the ellipse $\frac{x^{2}}{50} + \frac{y^{2}}{20} = 1$ from which a pair of perpendicular tangents can be drawn to the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$.

$A$ point on the curve $x=3 \cos \theta, y=2 \sin \theta$ at which the tangent is perpendicular to the $X$-axis is

Let a focus of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be $S(4, 0)$ and its eccentricity be $\frac{4}{5}$. If the point $P(3, \alpha)$ lies on $E$ and $O$ is the origin, then the area of $\triangle POS$ is equal to: (in $/ 5$)

The area of the region enclosed by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is . . . . . . square units. (in $\pi$)

If tangents are drawn from the point $(2 + 13 \cos \theta, 3 + 13 \sin \theta)$ to the ellipse $\frac{(x-2)^2}{25} + \frac{(y-3)^2}{144} = 1$,then the angle between them is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module