Tangent is drawn at the point `(x_(i), y_(i))` on the curve y = f(x), which intersects the x-axis at `(x_(i+1), 0)`. Now again tangent is drawn at `(x_(i+1), y_(i+1))` on the curve which intersects the x-axis at `(x_(i+2), 0)` and the process is repeated n times i.e. i = 1, 2, 3,......,n.
If `x_(1), x_(2), x_(3),...,x_(n)` form a geometric progression with common ratio equal to 2 and the curve passes through (1, 2), then the curve is

Options

A. circle
B. hyperbola
C. ellipse
D. parabola

Answer

B

Explanation

Again `x_(2) = x_(1) - (y_(1))/(((dy)/(dx)))`
`because" "x_(1), x_(2), x_(3),.....x_(n)` are in G.P.
`" "(x_(2))/(x_(1))=1-(y_(1))/(x_(1)((dy)/(dx)))`
`rArr" "1-(y)/(x)(dx)/(dy)=2`
`rArr" "-(y)/(x)(dx)/(dy)=1`
`rArr" "(dx)/(x)=-(dy)/(y)`
Integrating, `log x = - log y + log c`
`rArr" "xy = c`
Curve passing through `(1, 2)" "rArr" "c = 2`
`therefore" "`Curve is xy = 2.