OA and OB are two mutually perpendicular chords of y2 = 4ax, 'O' being the origin. Line AB will always pass through the point

Question

OA and OB are two mutually perpendicular chords of y2 = 4ax, 'O' being the origin. Line AB will always pass through the point (A) (2a, 0) (B) (6a, 0) (C) (8a, 0) (D) (4a, 0)

Tardigrade Admin · Accepted Answer

Let A ≡ (at1 2 , 2at1 ), B ≡ (at2 2 , 2at2 ) Thus t1 t2 = - 4 Equation of line AB is 2x - (t1 + t2 )y + 2at1 t2 = 0 2x - (t1 + t2 )y - 8a = 0 2(x - 4a) +λ y = 0 Hence line AB passes through a fixed point (4a, 0)

Q. $O A$ and $OB$ are two mutually perpendicular chords of $y^{2} = 4 a x$ , ' $O$ ' being the origin. Line $A B$ will always pass through the point

(2a, 0)

(6a, 0)

(8a, 0)

(4a, 0)

Let $A \equiv (a t_{1}^{2}, 2 a t_{1}), B \equiv (a t_{2}^{2}, 2 a t_{2})$
Thus $t_{1} t_{2} = - 4$
Equation of line $A B$ is $2 x - (t_{1} + t_{2}) y + 2 a t_{1} t_{2} = 0$
$2 x - (t_{1} + t_{2}) y - 8 a = 0$
$2 (x - 4 a) + λ y = 0$
Hence line $A B$ passes through a fixed point $(4 a, 0)$

Q. OA and OB are two mutually perpendicular chords of y2=4ax, 'O' being the origin. Line AB will always pass through the point