SLOPE of parallel lines are equalslope of tangent is defined as dy/dxCalculation:Given hyperbola is x2 - y2 = 3We know that slope of tangent is defined as dy/dxDifferentiating above equation WRT. x,2x - 2y.dy/dx = 0⇒ dy/dx = 2x/2y = x/y∴ Slope = m1 = dy/dx = x/yNow, Equation of straight line 2x + y + 8 = 0 Differentiating above equation with respect to x,2 + dy/dx = 0⇒ - 2 = dy/dx∴ slope = m2 = dy/dx = - 2 We know that the slope of parallel lines are equal⇒ m1 = m2⇒ x/y = - 2⇒ x = - 2y ---- (i)Now PUT this value in the equation of the hyperbolax2 - y2 = 3⇒ (- 2y)2 - y2 = 3⇒ 4y2 - y2 = 3⇒ 3y2 = 3⇒ y = ± 1NOW, at y = +1x = - 2 (Using equation (i))And, at y = - 1x = 2 (Using equation (i))∴ Points will be: (2, -1) or (-2, 1)

"> SLOPE of parallel lines are equalslope of tangent is defined as dy/dxCalculation:Given hyperbola is x2 - y2 = 3We know that slope of tangent is defined as dy/dxDifferentiating above equation WRT. x,2x - 2y.dy/dx = 0⇒ dy/dx = 2x/2y = x/y∴ Slope = m1 = dy/dx = x/yNow, Equation of straight line 2x + y + 8 = 0 Differentiating above equation with respect to x,2 + dy/dx = 0⇒ - 2 = dy/dx∴ slope = m2 = dy/dx = - 2 We know that the slope of parallel lines are equal⇒ m1 = m2⇒ x/y = - 2⇒ x = - 2y ---- (i)Now PUT this value in the equation of the hyperbolax2 - y2 = 3⇒ (- 2y)2 - y2 = 3⇒ 4y2 - y2 = 3⇒ 3y2 = 3⇒ y = ± 1NOW, at y = +1x = - 2 (Using equation (i))And, at y = - 1x = 2 (Using equation (i))∴ Points will be: (2, -1) or (-2, 1)

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The tangents to the hyperbola x2 - y2 = 3 are parallel to the straight line 2x + y + 8 = 0 at the following points

General Knowledge General Awareness in General Knowledge 8 months ago

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Concept:SLOPE of parallel lines are equalslope of tangent is defined as dy/dxCalculation:Given hyperbola is x2 - y2 = 3We know that slope of tangent is defined as dy/dxDifferentiating above equation WRT. x,2x - 2y.dy/dx = 0⇒ dy/dx = 2x/2y = x/y∴ Slope = m1 = dy/dx = x/yNow, Equation of straight line 2x + y + 8 = 0 Differentiating above equation with respect to x,2 + dy/dx = 0⇒ - 2 = dy/dx∴ slope = m2 = dy/dx = - 2 We know that the slope of parallel lines are equal⇒ m1 = m2⇒ x/y = - 2⇒ x = - 2y ---- (i)Now PUT this value in the equation of the hyperbolax2 - y2 = 3⇒ (- 2y)2 - y2 = 3⇒ 4y2 - y2 = 3⇒ 3y2 = 3⇒ y = ± 1NOW, at y = +1x = - 2 (Using equation (i))And, at y = - 1x = 2 (Using equation (i))∴ Points will be: (2, -1) or (-2, 1)

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Posted on 19 Nov 2024, this text provides information on General Knowledge related to General Awareness in General Knowledge. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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