P is a point on the hyperbola frac{x^2}{a^2} | vedclass

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(B) Let the point $P$ be $(a \sec 	heta, b 	an 	heta)$.Since $N$ is the foot of the perpendicular from $P$ on the transverse axis (x-axis),the coor

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$a^2$

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(B) Let the point $P$ be $(a \sec 	heta, b 	an 	heta)$.Since $N$ is the foot of the perpendicular from $P$ on the transverse axis (x-axis),the coordinates of $N$ are $(a \sec 	heta, 0)$. Thus,$ON = a \sec 	heta$.The equation of the tangent at $P(a \sec 	heta, b 	an 	heta)$ is $\frac{x \sec 	heta}{a} - \frac{y 	an 	heta}{b} = 1$.To find the x-intercept $T$,set $y = 0$:$\frac{x \sec 	heta}{a} = 1 \implies x = a \cos 	heta$.Thus,$OT = a \cos 	heta$.Therefore,$OT \cdot ON = (a \cos 	heta) \cdot (a \sec 	heta) = a^2 \cdot 1 = a^2$.

$P$ is a point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. $N$ is the foot of the perpendicular from $P$ on the transverse axis. The tangent to the hyperbola at $P$ meets the transverse axis at $T$. If $O$ is the centre of the hyperbola,then $OT \cdot ON$ is equal to:

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