If the tangent to y^{2}=4ax at the point (at^ | vedclass

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(A) The equation of the tangent to the parabola $y^{2}=4ax$ at the point $(at^{2}, 2at)$ is given by $ty = x + at^{2}$, which can be rewritten as $x -

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$t=-\operatorname{cosec} 	heta$

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(A) The equation of the tangent to the parabola $y^{2}=4ax$ at the point $(at^{2}, 2at)$ is given by $ty = x + at^{2}$, which can be rewritten as $x - ty + at^{2} = 0$. The equation of the normal to the hyperbola $x^{2}-y^{2}=a^{2}$ at the point $(a \sec 	heta, a 	an 	heta)$ is given by $\frac{ax}{\sec 	heta} + \frac{ay}{	an 	heta} = a^{2} + a^{2}$, which simplifies to $x \cos 	heta + y \cot 	heta = 2a$. Comparing the two equations $x - ty + at^{2} = 0$ and $x \cos 	heta + y \cot 	heta - 2a = 0$, we have the ratios of coefficients: $\frac{1}{\cos 	heta} = \frac{-t}{\cot 	heta} = \frac{at^{2}}{-2a}$. From $\frac{1}{\cos 	heta} = \frac{-t}{\cot 	heta}$, we get $t = -\frac{\cot 	heta}{\cos 	heta} = -\operatorname{cosec} 	heta$. From $\frac{1}{\cos 	heta} = \frac{at^{2}}{-2a}$, we get $t^{2} = -2 \sec 	heta$.

If the tangent to $y^{2}=4ax$ at the point $(at^{2}, 2at)$ where $|t|>1$ is a normal to $x^{2}-y^{2}=a^{2}$ at the point $(a \sec \theta, a \tan \theta)$, then

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