A conic passes through the point (2, 4) and i | vedclass

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(C) Let the point of tangency be $(x, y)$ and the slope of the tangent be $m = \frac{dy}{dx}$.The equation of the tangent is $Y - y = m(X - x)$.The $X

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$(4, 4)$ and $(-4, -4)$

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(C) Let the point of tangency be $(x, y)$ and the slope of the tangent be $m = \frac{dy}{dx}$.The equation of the tangent is $Y - y = m(X - x)$.The $X$-intercept (where $Y=0$) is $X = x - \frac{y}{m}$.The $Y$-intercept (where $X=0$) is $Y = y - mx$.Since the point $(x, y)$ bisects the segment between the axes,we have $x = \frac{1}{2}(x - \frac{y}{m} + 0)$ $\Rightarrow 2x = x - \frac{y}{m}$ $\Rightarrow \frac{y}{m} = -x$.Substituting $m = \frac{dy}{dx}$,we get $y \frac{dx}{dy} = -x \Rightarrow \frac{dy}{y} = -\frac{dx}{x}$.Integrating both sides,$\ln y = -\ln x + C \Rightarrow xy = c$.Since the conic passes through $(2, 4)$,$c = 2 	imes 4 = 8$. Thus,the equation is $xy = 8$.This is a rectangular hyperbola $xy = c^2/2$ with $c^2/2 = 8 \Rightarrow c^2 = 16$.The standard form is $X'Y' = a^2/2$ where $X', Y'$ are axes rotated by $45^\circ$.The foci of $xy = c^2/2$ are $(\pm c, \pm c)$.Here $c^2/2 = 8$ $\Rightarrow c^2 = 16$ $\Rightarrow c = 4$.Thus,the foci are $(4, 4)$ and $(-4, -4)$.

$A$ conic passes through the point $(2, 4)$ and is such that the segment of any of its tangents at any point contained between the coordinate axes is bisected at the point of tangency. Then the foci of the conic are:

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