Tangent to hyperbola xy=c^2 at point P intersects the x -axis at T and the y -axis at T^ . Normal hyperbola at P intersects the x -axis at N and the y -axis at N^ . If the area of the triangles PNT a PN'T' are Δ and Δ^

Correct answer: B. 2 c^2 Equation of tangent x+t^2 y=2 c t T (2 c t, 0) and T^ ( 0, (2 c)/(t)) Equation of normal t^3 x- ty=c (t^4-1) ; N [c(t- 1 t^3 ), 0] ; N [ 0, c( 1 t -t^3)] aligned and N T= c t^3 +ct; N^ T^ =ct^3+ c t and Δ=(1)/(2)| c t^3 +ct| c t (1)/(Δ)= 2 t^4 c^2(1+t^4) and Δ^ =(1)/(2)|ct^3+ c t | ct and 1 Δ^ = 2 c^2(1+t^4) (1)/(Δ)+ 1 Δ^ = 2 c^2 aligned

JEE Advanced

Exam: JEE Advanced
Subject: Mathematics
Chapter: Hyperbola

JEE Advanced · Mathematics · Hyperbola

Tangent to hyperbola xy=c^2 at point P intersects the x -axis at T...

Tangent to hyperbola

\mathrm{xy}=\mathrm{c}^2

at point P intersects the x -axis at T and the y -axis at

\mathrm{T}^{\prime}

. Normal hyperbola at P intersects the x -axis at N and the

y

-axis at

\mathrm{N}^{\prime}

. If the area of the triangles PNT a PN'T' are

\Delta

and

\Delta^{\prime}

respectively then

\frac{1}{\Delta}+\frac{1}{\Delta^{\prime}}

, is equal to

Options

A $c^2$
B $\frac{2}{\mathrm{c}^2}$
C $\frac{1}{\mathrm{c}^2}$
D $\frac{\mathrm{c}^2}{2}$

Correct answer

B. $\frac{2}{\mathrm{c}^2}$

Step-by-step solution

Equation of tangent

x+t^2 y=2 c t \Rightarrow T \equiv(2 c t, 0)

and

\mathrm{T}^{\prime} \equiv \left(

\frac{2 c}{t}\right) Equation of normal \mathrm{t}^3

\mathrm{x}-

\mathrm{ty}=\mathrm{c}

\left(\mathrm{t}^4-1\right) ;

\mathrm{N} \equiv\left[\mathrm{c}\left(\mathrm{t}-\frac{1}{\mathrm{t}^3}\right),

0\right] ;

\mathrm{N} \equiv\left[

0, c\left(\frac{1}{\mathrm{t}}-\mathrm{t}^3\right)\right]

\begin{aligned} \text{ and } \Rightarrow N T=\frac{\mathrm{c}}{\mathrm{t}^3}+\mathrm{ct}; \mathrm{N}^{\prime} \mathrm{T}^{\prime}=\mathrm{ct}^3+\frac{\mathrm{c}}{\mathrm{t}}\\ \text{ and } \Rightarrow \Delta=\frac{1}{2}\left|\frac{\mathrm{c}}{\mathrm{t}^3}+\mathrm{ct}\right| \frac{\mathrm{c}}{\mathrm{t}} \Rightarrow \frac{1}{\Delta}=\frac{2 \mathrm{t}^4}{\mathrm{c}^2\left(1+\mathrm{t}^4\right)} \text { and } \Delta^{\prime}=\frac{1}{2}\left|\mathrm{ct}^3+\frac{\mathrm{c}}{\mathrm{t}}\right| \mathrm{ct}\\ \text{ and } \Rightarrow \frac{1}{\Delta^{\prime}}=\frac{2}{\mathrm{c}^2\left(1+\mathrm{t}^4\right)} \Rightarrow \frac{1}{\Delta}+\frac{1}{\Delta^{\prime}}=\frac{2}{\mathrm{c}^2} \end{aligned}

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