Question bank/JEE Main/Mathematics

Let PQ and RS be 2 perpendicular focal chords of a rectangular
hyperbola, which are not parallel to its axes, then
Ekkuk PQ o RS,d nks yEcor tks
bl

APQ = RSBPQ^2 + (RS)^2 = (latus rectum)^2 (ukfHkyEc)^2CPQ + RS = (latus rectum) (ukfHkyEc)DNone of these buesa

Show answer and solution▾

[IMAGE] Let the parameteric equation of chord be x = a $\sqrt{2} + {rcos}\theta$
$\Rightarrow$ y = r sinθ
Solving it with x^2 - y^2 = a^2
We get $r^{2}(cos2\theta) + (2\sqrt{2}acos\theta)r + a^{2}$ = 0
(PQ)^2 = (r_1 + r_2 )^2 - 4r_1 r_2 =
$\frac{8a^{2}\cos^{2}\theta - 4a^{2}\cos{2\theta}}{\cos^{2}{2\theta}}$ =
4a^2 sec^22θ
Hence (RS)^2 = 4a^2
sec^2$\left\{ 2\left( \frac{\pi}{2} + \theta \right) \right\}$ = 4a^2
sec^22θ
Hindi
[IMAGE] Ekkuk thok dk izkpy lehdj.k x = a $\sqrt{2} + {rcos}\theta$
$\Rightarrow$ y = r sinθ gSA
gy djus ij x^2 - y^2 = a^2ds lkFk
gesa izkIr gksrk  $r^{2}(cos2\theta) + (2\sqrt{2}acos\theta)r + a^{2}$
= 0
(PQ)^2 = (r_1 + r_2 )^2 - 4r_1 r_2 =
$\frac{8a^{2}\cos^{2}\theta - 4a^{2}\cos{2\theta}}{\cos^{2}{2\theta}}$ =
4a^2 sec^22θ
vr (RS)^2 = 4a^2
sec^2$\left\{ 2\left( \frac{\pi}{2} + \theta \right) \right\}$ = 4a^2
sec^22θ

Source: Grass-G / M-27 Mathematics Paper 2 15 Jan.docx