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  • 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 co-ordinate axes is bisected at the point of tangency. Then the foci of

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 co-ordinate axes is bisected at the point of tangency. Then the foci of the conic are:

Option: 1

\mathrm{(2 \sqrt{2}, 0)\, \, \&\, (-2 \sqrt{2}, 0)}


Option: 2

\mathrm{(2 \sqrt{2}, 2 \sqrt{2}) \, \, \&\, \, (-2 \sqrt{2},-2 \sqrt{2})}


Option: 3

\mathrm{(4,4) \, \&\, (-4,-4)}


Option: 4

\mathrm{(4 \sqrt{2}, 4 \sqrt{2})\, \&\, (-4 \sqrt{2},-4 \sqrt{2})}


Answers (1)

best_answer

\mathrm{\begin{aligned} & T: Y-y=m(X-x) \\ & X=0, Y=y-m x \end{aligned}}

\mathrm{\mathrm{Y}=0, \mathrm{X}=\mathrm{x}-\frac{\mathrm{y}}{\mathrm{m}}}

\mathrm{\therefore \quad x-\frac{y}{m}=2 x \quad \Rightarrow \quad \frac{y}{m}=-x}

\mathrm{\frac{d y}{d x}=-\frac{y}{x} \Rightarrow \frac{d y}{y}+\frac{d x}{x}=0 \Rightarrow \ln x y=c \Rightarrow x y=c}

\mathrm{x=2 \, \& \, y=4 \Rightarrow equaiton\, is \, \, x y=8\, \, solving \, with \, \mathrm{y}=\mathrm{x}}

\mathrm{\begin{aligned} & \quad \mathrm{x}=2 \sqrt{2} \\ & \therefore \quad \mathrm{OA}=\sqrt{8+8}=4 \\ & \mathrm{OS}=4 \sqrt{2} \end{aligned}}

\mathrm{\text { coordinates of S are }(4,4) \text { or }(-4,-4) \Rightarrow(C)}

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vishal kumar

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