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\mathrm{P}  is any point on the hyperbola   \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}, tangent drawn to the hyperbola at \mathrm{P}  meets the  \mathrm{x}-axis at \mathrm{A}. If \mathrm{B}  is the foot of perpendicular drawn from \mathrm{P}  to the \mathrm{x} -axis, then OA.OB
(\mathrm{O} being the origin), is equal to
 

Option: 1

\mathrm{a}^2


Option: 2

\mathrm{b}^2


Option: 3

a  b


Option: 4

\frac{a b}{2}


Answers (1)

Let  \mathrm{ P \equiv(a \sec \theta, b \tan \theta)}


Equation of tangent at \mathrm{P}  is


\frac{x}{a} \sec \theta-\frac{y}{b} \tan \theta=1 \\


 Thus,  A \equiv(\operatorname{acos} \theta, 0), B \equiv(\operatorname{asec} \theta, 0) \\ 


\Rightarrow \text { OA.OB }=\operatorname{acos} \theta \cdot \operatorname{asec} \theta=a^2

Posted by

Ramraj Saini

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