"Let A (sec$\theta$, 2tan$\theta$) and B (sec$\phi$, 2tan$\phi$), where $\theta$ + $\phi$ = $\pi$/2, be two points on the hyperbola 2x2 $-$ y2 = 2. If ($\alpha$, $\beta$) is the point of the intersection of the normals to the hyperbola at A and B, then (2$\beta$)2 is equal to ____________."

Question

Accepted Answer

"Since, point A (sec$\theta$, 2tan$\theta$) lies on the hyperbola 2x² $-$ y² = 2

Therefore, 2sec²$\theta$ $-$ 4tan²$\theta$ = 2

$\Rightarrow$ 2 + 2tan²$\theta$ $-$ 4tan²$\theta$ = 2

$\Rightarrow$ tan$\theta$ = 0 $\Rightarrow$ $\theta$ = 0

Similarly, for point B, we will get $\phi$ = 0.

but according to question $\theta$ + $\phi$ = ${\pi \over 2}$ which is not possible.

Hence, it must be a 'BONUS'."

Let A (sec$\theta$, 2tan$\theta$) and B (sec$\phi$, 2tan$\phi$), where $\theta$ + $\phi$ = $\pi$/2, be two points on the hyperbola 2x² $-$ y² = 2. If ($\alpha$, $\beta$) is the point of the intersection of the normals to the hyperbola at A and B, then (2$\beta$)² is equal to ____________.

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Let A (sec$\theta$, 2tan$\theta$) and B (sec$\phi$, 2tan$\phi$), where $\theta$ + $\phi$ = $\pi$/2, be two points on the hyperbola 2x2 $-$ y2 = 2. If ($\alpha$, $\beta$) is the point of the intersection of the normals to the hyperbola at A and B, then (2$\beta$)2 is equal to ____________.

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Let A (sec$\theta$, 2tan$\theta$) and B (sec$\phi$, 2tan$\phi$), where $\theta$ + $\phi$ = $\pi$/2, be two points on the hyperbola 2x² $-$ y² = 2. If ($\alpha$, $\beta$) is the point of the intersection of the normals to the hyperbola at A and B, then (2$\beta$)² is equal to ____________.