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The tangent to the hyperbola \mathrm{x y=c^2} at the point P intersects the x-axis at T and the y-axis at \mathrm{T^{\prime}.} The normal to the hyperbola at \mathrm{\mathrm{P}} intersects the \mathrm{\mathrm{x}}-axis at \mathrm{\mathrm{N}} and the \mathrm{\mathrm{y}}-axis at \mathrm{\mathrm{N}^{\prime}}. The areas of the triangles PNT and PN'T' are \mathrm{\Delta} and \mathrm{\Delta^{\prime}} respectively, then \mathrm{\frac{1}{\Delta}+\frac{1}{\Delta^{\prime}}} is

Option: 1

equal to  1


Option: 2

depends on  t


Option: 3

depends on c


Option: 4

equal to 2


Answers (1)

best_answer

\mathrm{\text { Tangent: } \frac{\mathrm{x}}{\mathrm{ct}}+\frac{\mathrm{yt}}{\mathrm{c}}=2}

\mathrm{\text { put } \quad y=0 \quad x=2 \text { ct }(T)}

     \mathrm{x=0 \quad y=\frac{2 c}{t}(T)}

\mathrm{\text { normal is } y-\frac{c}{t}=t^2(x-c t)}

\mathrm{\text { put } \quad y=0 \quad x=c t-\frac{c}{t^3}(N)}

\mathrm{\mathrm{x}=0 \quad \frac{\mathrm{c}}{\mathrm{t}}-\mathrm{ct}^3 \quad\left(\mathrm{~N}^{\prime}\right)}

\mathrm{\text { Area of } \Delta \text { PNT }=\frac{c}{2 t}\left(c t+\frac{c}{t^3}\right) \quad \Rightarrow \quad \Delta=\frac{c^2\left(1+t^4\right)}{2 t^4}}

\mathrm{\text { area of } \Delta \mathrm{PN}^{\prime} \mathrm{T}^{\prime}=\mathrm{ct}\left(\frac{\mathrm{c}}{\mathrm{t}}+\mathrm{ct}^3\right) \quad \Rightarrow \quad \Delta^{\prime}=\frac{\mathrm{c}^2\left(1+\mathrm{t}^4\right)}{2}}

\mathrm{\therefore \quad \frac{1}{\Delta}+\frac{1}{\Delta^{\prime}}=\frac{2 \mathrm{t}^4}{\mathrm{c}^2\left(1+\mathrm{t}^4\right)}+\frac{2}{\mathrm{c}^2\left(1+\mathrm{t}^4\right)}=\frac{2}{\mathrm{c}^2\left(1+\mathrm{t}^4\right)}\left(\mathrm{t}^4+1\right)=\frac{2}{\mathrm{c}^2}}

which is independent of t.

Posted by

Devendra Khairwa

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