At any point (x, y) on a curve,if the length | vedclass

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(D) The length of the subnormal at any point $(x, y)$ is given by $y \frac{dy}{dx}$.Given $y \frac{dy}{dx} = x - 1$.Integrating both sides with respec

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$(1, 2)$

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(D) The length of the subnormal at any point $(x, y)$ is given by $y \frac{dy}{dx}$.Given $y \frac{dy}{dx} = x - 1$.Integrating both sides with respect to $x$:$\int y \, dy = \int (x - 1) \, dx$$\frac{y^2}{2} = \frac{x^2}{2} - x + C$$y^2 = x^2 - 2x + 2C$.Since the curve passes through $(1, 2)$,we substitute $x = 1$ and $y = 2$:$2^2 = 1^2 - 2(1) + 2C$$4 = 1 - 2 + 2C$$4 = -1 + 2C \implies 2C = 5$.Substituting $2C = 5$ into the equation:$y^2 = x^2 - 2x + 5$$y^2 - (x^2 - 2x + 1) = 4$$y^2 - (x - 1)^2 = 4$Dividing by $4$:$\frac{y^2}{4} - \frac{(x - 1)^2}{4} = 1$.This is a hyperbola with center $(1, 0)$.The vertices of the hyperbola $\frac{y^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$ are $(h, k \pm a)$.Here $h = 1, k = 0, a = 2$.Vertices are $(1, 0 \pm 2)$,i.e.,$(1, 2)$ and $(1, -2)$.Comparing with the options,$(1, 2)$ is a vertex.

At any point $(x, y)$ on a curve,if the length of the subnormal is $(x - 1)$ and the curve passes through $(1, 2)$,then the curve is a conic. $A$ vertex of the curve is:

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