JEE Main 2022 (Online) 25th July Morning ShiftQuestion 1
Let the equation of two diameters of a circle $x^2+y^2-2x+2fy+1=0$ be $2px-y=1$ and $2x+py=4p$. Then the slope $m\in(0,\infty)$ of the tangent to the hyperbola $3x^2-y^2=3$ passing through the centre of the circle is equal to ________.
Hyperbola
Answer$2$
Solution
Centre $(1,-f)$ lies on both diameters: $2p+f=1$ and $2-pf=4p$, so $2p^2-5p+2=0$, $p=2$ or $\frac12$. For $p=\frac12$, $f=0$ and the circle $(x-1)^2+y^2=0$ is degenerate, so $p=2$, $f=-3$, centre $(1,3)$. Tangent $y=mx+c$ to $x^2-\frac{y^2}3=1$: $c^2=m^2-3$; through $(1,3)$: $c=3-m$, so $(3-m)^2=m^2-3\Rightarrow m=2$.