Assertion (A): The curves y^2 = 4x and x^2 = | vedclass

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(D) For the curve $y^2 = 4x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} = \frac{2}{y}$. At $(2, -2)$,$m_1 = \f

Vedclass · Accepted Answer

$(A)$ is false but $(R)$ is true

Vedclass · Answer

(D) For the curve $y^2 = 4x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} = \frac{2}{y}$. At $(2, -2)$,$m_1 = \frac{2}{-2} = -1$.For the curve $x^2 = -2y$,differentiating with respect to $x$ gives $2x = -2 \frac{dy}{dx}$,so $\frac{dy}{dx} = -x$. At $(2, -2)$,$m_2 = -2$.The product of slopes $m_1 	imes m_2 = (-1) 	imes (-2) = 2 
eq -1$. Thus,the curves do not intersect orthogonally at $(2, -2)$.At the origin $(0,0)$,the tangents are $x=0$ (vertical) and $y=0$ (horizontal),which are perpendicular,but the assertion claims intersection at $(1,2)$ which is incorrect as $(1,2)$ does not lie on the curves.Therefore,Assertion $(A)$ is false and Reason $(R)$ is true.

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