The condition that the two curves y^2 = 4ax a | vedclass

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(D) Let the point of intersection be $(x_1, y_1)$.For the curve $y^2 = 4ax$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4a$,so $\fra

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$c^4 = 32a^4$

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(D) Let the point of intersection be $(x_1, y_1)$.For the curve $y^2 = 4ax$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4a$,so $\frac{dy}{dx} = \frac{2a}{y}$.Thus,the slope of the tangent at $(x_1, y_1)$ is $m_1 = \frac{2a}{y_1}$.For the curve $xy = c^2$,differentiating with respect to $x$ gives $x \frac{dy}{dx} + y = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.Thus,the slope of the tangent at $(x_1, y_1)$ is $m_2 = -\frac{y_1}{x_1}$.Since the curves cut orthogonally,$m_1 	imes m_2 = -1$.$\frac{2a}{y_1} 	imes (-\frac{y_1}{x_1}) = -1 \Rightarrow \frac{2a}{x_1} = 1 \Rightarrow x_1 = 2a$.Since $(x_1, y_1)$ lies on $y^2 = 4ax$,$y_1^2 = 4a(2a) = 8a^2$.Since $(x_1, y_1)$ lies on $xy = c^2$,$x_1 y_1 = c^2$,so $(x_1 y_1)^2 = c^4$.Substituting $x_1 = 2a$ and $y_1^2 = 8a^2$,we get $c^4 = x_1^2 y_1^2 = (2a)^2 (8a^2) = 4a^2 	imes 8a^2 = 32a^4$.

The condition that the two curves $y^2 = 4ax$ and $xy = c^2$ cut orthogonally is

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