EllipseeasyFree

Ellipse: Take Four Points Following Correct

JEE Maths question with a full step-by-step solution.

Question

Take the four points

\left(\pm\frac{ab}{\sqrt{a^{2}+b^{2}}},\ \pm\frac{ab}{\sqrt{a^{2}+b^{2}}}\right)

Which of the following is correct?

AOnly two ellipses can be drawn through themcorrect

BOnly one ellipse can be drawn through them

CNo parabola can be drawn through themcorrect

DExactly two parabolas can be drawn through them

Solution

Step 1: Put

c = \frac{ab}{\sqrt{a^{2}+b^{2}}}

so the four points are

(\pm c, \pm c)

, the vertices of a square centred at the origin with sides parallel to the axes. Here

a \ne 0

and

b \ne 0

, so

c \ne 0

. Step 2: On

\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1

\frac{c^{2}}{a^{2}} + \frac{c^{2}}{b^{2}} = c^{2}\cdot\frac{a^{2}+b^{2}}{a^{2}b^{2}} = \frac{a^{2}b^{2}}{a^{2}+b^{2}}\cdot\frac{a^{2}+b^{2}}{a^{2}b^{2}} = 1 .

Only

c^{2}

enters, so all four points satisfy it. Step 3: The left side is symmetric in

a

and

b

, so the same computation puts the four points on

\dfrac{x^{2}}{b^{2}} + \dfrac{y^{2}}{a^{2}} = 1

as well. Two ellipses of this family pass through them, which is option (1), and option (2) is false. Step 4: For a parabola, the mid points of a system of parallel chords lie on a line parallel to its axis. Let a parabola pass through all four points. Then

x = c

and

x = -c

are two parallel chords of it, with mid points

(c,0)

and

(-c,0)

; these are distinct as

c \ne 0

, so the axis of the parabola is parallel to the

x

-axis. Step 5: Again,

y = c

and

y = -c

are two parallel chords, with distinct mid points

(0,c)

and

(0,-c)

, so the axis is parallel to the

y

-axis also, which is not possible. Hence no parabola passes through the four points: option (3) is true and option (4) is false. Answer: (1) and (3).

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