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A curve has the equation, 6x^2 +3xy−y^2 +6=0 and passes through the point A (-5, 10). Find the equation of the normal to the curve at A.

Use implicit differentiation on original equation-
12x + 3x(dy/dx) + 3y - 2y(dy/dx) = 0
dy/dx= -12x -3y/(3x-2y) at A, x= -5 and y= 10 therefore, dy/dx=-6/7

To find the normal of t...

CG
Answered by Chantal G. Maths tutor
5775 Views

A curve has the equation y = (x^2 - 5)e^(x^2). Find the x-coordinates of the stationary points of the curve.

This requires the chain rule and the product rule to be used to differentiate the function. The substitution u = x2 can be used to make this easier. Using this, du/dx = 2x and y = (u-5)eu<...

OJ
Answered by Oliver J. Maths tutor
3572 Views

Solve x^2 +11x +30 = 0

This is what we call a quadratic equation. To solve, first we need to factorise this equation into two brackets. (x + _ )(x + _) =0To fill the brackets, we need to find 2 numbers that add to 11 and multip...

HR
Answered by Holly R. Maths tutor
3903 Views

Find a local minimum of the function f(x) = x^3 - 2x.

To find a local minimum (i.e. a point where the function changes from a negative slope into a positive slope), we first need to find all points where the slope of the function is zero. The first derivativ...

KH
Answered by Karoline H. Maths tutor
3466 Views

The equation of line A is (x)^2 + 11x + 12 = y - 4, while the equation of line B is x - 6 = y + 2. Find the co-ordinate(s) of the point at which lines A and B intersect.

While this question may seem complicated, this question is simply asking you to solve the equations of these two lines as simultaneous equations. Line A: x2 + 11x + 12 = y - 4 --> x

AA
Answered by Ann A. Maths tutor
2830 Views

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