Chords of an ellipse frac{x^2}{a^2} + frac{y^ | vedclass

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

Question

(C) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The positive end of the minor axis is $P(0, b)$.Let $(h, k)$ be the midpoint of a chor

Vedclass · Accepted Answer

an ellipse

Vedclass · Answer

(C) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The positive end of the minor axis is $P(0, b)$.Let $(h, k)$ be the midpoint of a chord passing through $(0, b)$ and some point $(x_1, y_1)$ on the ellipse.Then, $h = \frac{x_1 + 0}{2} \Rightarrow x_1 = 2h$ and $k = \frac{y_1 + b}{2} \Rightarrow y_1 = 2k - b$.Since $(x_1, y_1)$ lies on the ellipse, we have $\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} = 1$.Substituting the values of $x_1$ and $y_1$, we get $\frac{(2h)^2}{a^2} + \frac{(2k - b)^2}{b^2} = 1$.This simplifies to $\frac{4h^2}{a^2} + \frac{4(k - b/2)^2}{b^2} = 1$, which can be rewritten as $\frac{h^2}{(a/2)^2} + \frac{(k - b/2)^2}{(b/2)^2} = 1$.Replacing $(h, k)$ with $(x, y)$, the locus is $\frac{x^2}{(a/2)^2} + \frac{(y - b/2)^2}{(b/2)^2} = 1$, which represents an ellipse.

Chords of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are drawn through the positive end of the minor axis $(0, b)$. The locus of their midpoints lies on:

Similar Questions

The locus of a point which moves such that the sum of its distances from two fixed points is a constant,is

The equation of the normal at the point $(0, 3)$ of the ellipse $9x^2 + 5y^2 = 45$ is

If the chord of the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ having $(1,1)$ as its midpoint is $x+\alpha y=\beta$,then

If a normal is drawn at a variable point $P(x, y)$ on the curve $9x^2 + 16y^2 = 144$,then the maximum distance from the centre of the curve to the normal is

An ellipse having foci at $(3, 3)$ and $(-4, 4)$ and passing through the origin has eccentricity equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module