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Question

Let P(asecθ,btanθ) and Q(asecϕ,btanϕ), where θ+ϕ=π2, be two points on the hyperbola x2a2-y2b2=1. If (h,k) is the point of intersection of the normal at P and Q, thenk=


A

[a2+b2]a

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B

[a2+b2]a

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C

[a2+b2]b

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D

[a2+b2]b

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Solution

The correct option is D

[a2+b2]b


Explanation for the correct option:

Finding the value of k:

The slope of normal to x2a2-y2b2=1at (asecθ,btanθ)

2xa2-2yb2×dydx=0dydx=b2a2xy

Slope for normal at the point (asecθ,btanθ) will be [a2btanθ]b2asecθ=absinθ

∴ Equation of normal at(asecθ,btanθ) is ybtanθ=absinθ(xasecθ)

(asinθ)x+by=(a2+b2)tanθax+bcosecθ=(a2+b2)secθ..(i)

Similarly, the equation of normal to x2a2-y2b2=1at (asecϕ,btanϕ), y=(a2+b2)(secθsecϕ).(ii)

On subtracting equation (ii) from equation(i), b(cosecθcosecϕ)y=(a2+b2)(secθsecϕ)

y=(a2+b2)b.secθsecϕcosecθcosecϕ=(a2+b2)b.secθsecπ2-θcscθcscπ2-θ[ϕ+θ=π2]=(a2+b2)b.1y=[a2+b2]b

Therefore, the correct answer is option (D).


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