Match the points on the curve 2y^2 = x + 1 wi | vedclass

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(A) The given equation of the curve is $2y^2 = x + 1$.Differentiating both sides with respect to $x$,we get:$4y \frac{dy}{dx} = 1 \Rightarrow \frac{dy

Vedclass · Accepted Answer

$A-2, B-5, C-3, D-1$

Vedclass · Answer

(A) The given equation of the curve is $2y^2 = x + 1$.Differentiating both sides with respect to $x$,we get:$4y \frac{dy}{dx} = 1 \Rightarrow \frac{dy}{dx} = \frac{1}{4y}$.The slope of the tangent at any point $(x, y)$ is $m_t = \frac{1}{4y}$.The slope of the normal at any point $(x, y)$ is $m_n = -\frac{1}{m_t} = -4y$.Now,we calculate the slope of the normal at each given point:$A. (7, 2): m_n = -4(2) = -8$ (Matches $2$).$B. (0, 1/\sqrt{2}): m_n = -4(1/\sqrt{2}) = -2\sqrt{2}$ (Matches $5$).$C. (1, -1): m_n = -4(-1) = 4$ (Matches $3$).$D. (3, \sqrt{2}): m_n = -4(\sqrt{2}) = -4\sqrt{2}$ (Matches $1$).Thus,the correct matching is $A-2, B-5, C-3, D-1$.

Point	Slope of the normal
$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

Match the points on the curve $2y^2 = x + 1$ with the slopes of the normals at those points and choose the correct answer.
Point Slope of the normal
$A. (7, 2)$ $1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$ $2. -8$
$C. (1, -1)$ $3. 4$
$D. (3, \sqrt{2})$ $4. 0$
$5. -2\sqrt{2}$

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Match the points on the curve $2y^2 = x + 1$ with the slopes of the normals at those points and choose the correct answer.PointSlope of the normal$A. (7, 2)$$1. -4\sqrt{2}$$B. (0, 1/\sqrt{2})$$2. -8$$C. (1, -1)$$3. 4$$D. (3, \sqrt{2})$$4. 0$$5. -2\sqrt{2}$