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

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

Question

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

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

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

Similar Questions

If the length of the tangent drawn from the point $(5, 3)$ to the circle $x^2 + y^2 + ky + 17 = 0$ is $7$,then $k = \dots$

For what possible value of $p$ is the line $x \cos \alpha + y \sin \alpha = p$ a tangent to the circle $x^2 + y^2 - 2qx \cos \alpha - 2qy \sin \alpha = 0$?

The slope of the common tangent drawn to the circles $x^2+y^2-4x+12y-216=0$ and $x^2+y^2+6x-12y+36=0$ is

If $3x + y + k = 0$ is a tangent to the circle $x^2 + y^2 = 10$,then $k = . . . . . . $.

The equation of a normal to the circle $x^2+y^2-2x=0$ that is parallel to the line $x+2y-3=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Match the points on the curve $2y^2 = x + 1$ with the slopes of the normals at those points and choose the correct answer.$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}$