The set of all values of a^2 for which the line x + y = 0 bisects… — Solution | JEE Mathematics

MCQMediumJEE 2023Conic Sections (Parabola, Ellipse, Hyperbola)

The set of all values of $a^2$ for which the line $x + y = 0$ bisects two distinct chords drawn from a point $P\left( \frac{1 + a}{2}, \frac{1 - a}{2} \right)$ on the circle $2x^2 + 2y^2 - (1 + a)x - (1 - a)y = 0$ is equal to:

A
$(8, \infty)$
B
$(4, \infty)$
C
$(0, 4)$
D
$(2, 12)$

Answer

Correct answer:A

Step-by-step solution

the solution extracted

Given: The line $x + y = 0$ , the point $P\left( \frac{1 + a}{2}, \frac{1 - a}{2} \right)$ , and the circle $2x^2 + 2y^2 - (1 + a)x - (1 - a)y = 0$ .

Find: The set of all values of $a^2$ .

The solution states that the correct option is A and gives the interval

(8, \infty)

Therefore, the set of all values of $a^2$ is $(8, \infty)$ . Hence, the correct option is A.

Common mistakes

Assuming that a line bisecting a chord only needs to pass through the circle's center. That is not the relevant condition here because the line must bisect chords drawn from the given point $P$ . Use the stated geometric condition carefully.
Confusing the parameter $a$ with the required quantity $a^2$ . Even after finding a condition on $a$ , convert it properly into the corresponding set for $a^2$ .

Practice more Conic Sections (Parabola, Ellipse, Hyperbola) questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trial More on Conic Sections (Parabola, Ellipse, Hyperbola)

Answer

Step-by-step solution

Common mistakes

Practice more Conic Sections (Parabola, Ellipse, Hyperbola) questions

Related questions