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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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.

Similar Questions

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

Which of the following represents a parabola?

The line $y = 2x + c$ is tangent to the parabola $y^2 = 4x$,then $c = $

$A$ parabola having its axis parallel to the $Y$-axis passes through the points $(0, 2/5)$,$(4, -2)$,and $(1, 8/5)$. Which of the following points lies on this parabola?

$A$ point on the parabola whose focus is $S(1,-1)$ and whose vertex is $A(1,1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module