JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Hyperbola

The center of the circle $x^2+y^2-2x+2fy+1=0$ is $(1,-f)$.

Since both lines are diameters, they pass through the center:

• From $2px-y=1$: $2p(1)-(-f)=1\Rightarrow 2p+f=1$
• From $2x+py=4p$: $2(1)+p(-f)=4p\Rightarrow 2-pf=4p$

From the second equation: $2=p(f+4)$, so $p=\frac{2}{f+4}$.

Substituting into the first: $2\cdot \frac{2}{f+4}+f=1$

$\frac{4}{f+4}=1-f$

$4=(1-f)(f+4)=f+4-f^2-4f=4-3f-f^2$

$f^2+3f=0\Rightarrow f=0$ or $f=-3$.

For $f=0$: center is $(1,0)$; for $f=-3$: center is $(1,3)$.

The hyperbola is $x^2-\frac{y^2}{3}=1$. Tangent with slope $m$: $y=mx\pm \sqrt{m^2-3}$.

For center $(1,0)$: $0=m\pm \sqrt{m^2-3}$ gives no real positive $m$.

For center $(1,3)$: $3=m\pm \sqrt{m^2-3}$

Taking the minus sign: $(3-m{)}^2=m^2-3\Rightarrow 9-6m+m^2=m^2-3\Rightarrow 6m=12\Rightarrow m=2$.