Ellipse

We are given that

AB

is a variable chord of the ellipse

%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%20%2B%20%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%20%3D%201

, and it subtends a right angle at the origin

O

. Let the coordinates of

A

and

B

(x_1%2C%20y_1)

and

(x_2%2C%20y_2)

, respectively. The condition that the chord

AB

subtends a right angle at the origin

O

implies that the vectors

%5Coverrightarrow%7BOA%7D

and

%5Coverrightarrow%7BOB%7D

are perpendicular. This means:

%5Coverrightarrow%7BOA%7D%20%5Ccdot%20%5Coverrightarrow%7BOB%7D%20%3D%200

Using the dot product, we have:

x_1%20x_2%20%2B%20y_1%20y_2%20%3D%200%20%5Ctag%7B1%7D

Now, we will use the parametric form of the ellipse. The parametric equations of the ellipse are:

x_1%20%3D%20a%20%5Ccos%20%5Ctheta_1%2C%20%5Cquad%20y_1%20%3D%20b%20%5Csin%20%5Ctheta_1

x_2%20%3D%20a%20%5Ccos%20%5Ctheta_2%2C%20%5Cquad%20y_2%20%3D%20b%20%5Csin%20%5Ctheta_2

Substituting these into equation (1):

a%20%5Ccos%20%5Ctheta_1%20%5Ccdot%20a%20%5Ccos%20%5Ctheta_2%20%2B%20b%20%5Csin%20%5Ctheta_1%20%5Ccdot%20b%20%5Csin%20%5Ctheta_2%20%3D%200

a%5E2%20%5Ccos%20%5Ctheta_1%20%5Ccos%20%5Ctheta_2%20%2B%20b%5E2%20%5Csin%20%5Ctheta_1%20%5Csin%20%5Ctheta_2%20%3D%200

Using the trigonometric identity

%5Ccos(%5Ctheta_1%20-%20%5Ctheta_2)%20%3D%20%5Ccos%20%5Ctheta_1%20%5Ccos%20%5Ctheta_2%20%2B%20%5Csin%20%5Ctheta_1%20%5Csin%20%5Ctheta_2

, we get:

a%5E2%20%5Ccos(%5Ctheta_1%20-%20%5Ctheta_2)%20%3D%200

Thus:

%5Ccos(%5Ctheta_1%20-%20%5Ctheta_2)%20%3D%200

This implies that:

%5Ctheta_1%20-%20%5Ctheta_2%20%3D%20%5Cpm%20%5Cfrac%7B%5Cpi%7D%7B2%7D

So, the points

A

and

B

lie on the ellipse and subtend a right angle at the origin. Next, we want to find

%5Cfrac%7B1%7D%7BOA%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7BOB%5E2%7D

. The distances

OA

and

OB

are given by:

OA%5E2%20%3D%20x_1%5E2%20%2B%20y_1%5E2%20%3D%20a%5E2%20%5Ccos%5E2%20%5Ctheta_1%20%2B%20b%5E2%20%5Csin%5E2%20%5Ctheta_1

OB%5E2%20%3D%20x_2%5E2%20%2B%20y_2%5E2%20%3D%20a%5E2%20%5Ccos%5E2%20%5Ctheta_2%20%2B%20b%5E2%20%5Csin%5E2%20%5Ctheta_2

Now, using the fact that

%5Ctheta_1%20-%20%5Ctheta_2%20%3D%20%5Cpm%20%5Cfrac%7B%5Cpi%7D%7B2%7D

, we can use the identity:

%5Ccos%5E2%20%5Ctheta_1%20%2B%20%5Csin%5E2%20%5Ctheta_1%20%3D%201%20%5Cquad%20%5Ctext%7Band%7D%20%5Cquad%20%5Ccos%5E2%20%5Ctheta_2%20%2B%20%5Csin%5E2%20%5Ctheta_2%20%3D%201

and find that:

%5Cfrac%7B1%7D%7BOA%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7BOB%5E2%7D%20%3D%20%5Cfrac%7B1%7D%7Ba%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bb%5E2%7D

Final Answer:

%5Cboxed%7B%5Cfrac%7B1%7D%7Ba%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bb%5E2%7D%7D

Let the standard equation of the ellipse be: $%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%20%2B%20%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%20%3D%201$

Let the chord be drawn through the point $(0%2C%20b)$ , which is the positive end of the minor axis. Let this chord intersect the ellipse at points $A$ and $B$ , and let $M$ be the midpoint of chord $AB$ . Let the coordinates of the midpoint $M$ be $(h%2C%20k)$ . Since the chord passes through $(0%2C%20b)$ and midpoint is $(h%2C%20k)$ , the other end of the chord is at: $2h%2C%202k%20-%20b$ Now, since both endpoints of the chord lie on the ellipse, their coordinates must satisfy the ellipse equation. Substitute the second endpoint into the ellipse: $%5Cfrac%7B(2h)%5E2%7D%7Ba%5E2%7D%20%2B%20%5Cfrac%7B(2k%20-%20b)%5E2%7D%7Bb%5E2%7D%20%3D%201$ Simplifying: $%5Cfrac%7B4h%5E2%7D%7Ba%5E2%7D%20%2B%20%5Cfrac%7B(2k%20-%20b)%5E2%7D%7Bb%5E2%7D%20%3D%201$ This is the locus of the midpoint $(h%2C%20k)$ . Multiply through by appropriate factors: $%5Cfrac%7B4h%5E2%7D%7Ba%5E2%7D%20%2B%20%5Cfrac%7B(2k%20-%20b)%5E2%7D%7Bb%5E2%7D%20%3D%201$ This equation represents the locus of the midpoints of all such chords, and hence, the locus is: $%5Cboxed%7B%5Cfrac%7B4x%5E2%7D%7Ba%5E2%7D%20%2B%20%5Cfrac%7B(2y%20-%20b)%5E2%7D%7Bb%5E2%7D%20%3D%201%7D$ where $(x%2C%20y)$ are the coordinates of the midpoint.

So, the midpoint of all such chords lies on a curve which is an ellipse, shifted vertically.

High-Yield Trend

Chapter Questions
6 MCQs

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Ellipse

High-Yield Trend

Chapter Questions 6 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Chapter Questions
6 MCQs