Consider the hyperbola `H:x^2-y^2=1` and a circle S with centre `N(x_2,0)` Suppose that H and S touch each other at a point `(P(x_1,y_1)` with `x_1 > 1 and y_1 > 0` The common tangent to H and S at P intersects the x-axis at point M. If (l,m) is the centroid of the triangle `DeltaPMN` then the correct expression is (A) `(dl)/(dx_1)=1-1/(3x_1^2)` for `x_1 > 1` (B) `(dm)/(dx_1) =x_!/(3(sqrtx_1^2-1))) for x_1 > 1` (C) `(dl)/(dx_1)=1+1/(3x_1^2) for x_1 > 1` (D) `(dm)/(dy_1)=1/3 for y_1 > 0`

Options

A. `(dl)/(dx_(1))=1-(1)/(3x_(1)^(2)` for `x_(1) gt 1`
B. `(dm)/(dx_(1))=(x_(1))/(3(sqrt(x_(1)^(2)-1)))` for `x_(1) gt 1`
C. `(dl)/(dx_(1))=1+(1)/(3x_(1)^(2)` for `x_(1) gt 1`
D. `(dm)/(dy)=(1)/(3)` for `y_(1) gt 0`.

Explanation

The equation of tangent at `P(x_(1),y_(1))` to the hyperbola `H : x^(2)-y^(2)=1` is
`x x_(1)=y y_(1)=1`…….`(i)`
This intersects `x`-axis at `M(1//x_(1),0)` .
Line `(i)` is tangent to the circle `S` whose centre is at `N(x_(2),0)`.
`:. ("Slope of "i)xx("Slope of"NP)=-1`
`implies (x_(1))/(y_(1))xx(y_(1)-0)/(x_(1)-x_(2))=-1impliesx_(2)=2x_(1)`
So, the coordinates of `N` are `(2x_(1),0)`.
It is given that `(l,m)` is the centroid of `DeltaPMN` the coordinates of whose vertices are `P(x_(1),y_(1))`, `M(1//x_(1),0)` and `N(2x_(1),0)`.
`:.l=(1)/(3)(x_(1)+(1)/(x_(1))+2x_(1))` and `m=(y_(1))/(3)`
`impliesl=x_(1)+(1)/(3)x_(1)` and `m=(y_(1))/(3)`
`implies(dl)/(dx_(1))=1-(3)/(x_(1)^(2))` and `(dm)/(dy)=(1)/(3)`
Since `(x_(1),y_(1))` lies on the hyperbola `x^(2)-y^(2)=1`. Therefore,
`x_(1)^(2)-y_(1)^(2)=1`.
`impliesy_(1)=sqrt(x_(1)^(2)-1)` [` :' y_(1) gt 0`]
`implies (dy_(1))/(dx_(1))=(x_(1))/(sqrt(x_(1)^(2))-1)`
`:. (dm)/(dx_(1))=(dm)/(dy_(1))xx(dy_(1))/(dx_(1))=(1)/(3)(x_(1))/(sqrt(x_(1)^(2)-1))`
Hence, `(dl)/(dx_(1))=1-(3)/(x_(1)^(2))` for `x_(1) gt 0`, `(dm)/(dy_(1))-(1)/(3)` for `y_(1) gt 0`
and `(dm)/(dx_(1))=(x_(1))/(3(sqrt(x_(1)^(2)-1)))`.
So, options `(a)`, `(b)` and `(d)` are true