Find the equations of the normals of the curve x^2+2y^2=152 which are parallel to the line 2x-6y-7=0. Please provide clear workings, thank you!
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The line 2x - 6y - 7 = 0 can be rewritten as y = x/3 - 7/6, which has slope 1/3.
So, the equations of the normals of the curve x^2 + 2y^2 = 152 must have slope 1/3.
==> The slope of a tangent line equals -1/(1/3) = -3.
However by implicit differentiation, we have
2x + 4y dy/dx = 0
==> dy/dx = -x/(2y).
So, we need to find points where -x/(2y) = -3 ==> x = 6y.
Substitute this into the equation of the curve:
(6y)^2 + 2y^2 = 152 ==> y^2 = 4 ==> y = ±2.
Hence, the points of tangency are (x, y) = (±12, ±2).
Finally, the equations of the normal lines are
y - (±12) = (1/3)(x - (±12)).
I hope this helps!
So, the equations of the normals of the curve x^2 + 2y^2 = 152 must have slope 1/3.
==> The slope of a tangent line equals -1/(1/3) = -3.
However by implicit differentiation, we have
2x + 4y dy/dx = 0
==> dy/dx = -x/(2y).
So, we need to find points where -x/(2y) = -3 ==> x = 6y.
Substitute this into the equation of the curve:
(6y)^2 + 2y^2 = 152 ==> y^2 = 4 ==> y = ±2.
Hence, the points of tangency are (x, y) = (±12, ±2).
Finally, the equations of the normal lines are
y - (±12) = (1/3)(x - (±12)).
I hope this helps!
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It is simple especially if you use vectors for it
The line 2x - 6y - 7 = 0 has slope 1/3 and a vector along it is <3,1>.
The normals to the curve are parallel to the gradient of the function G(x) = x^2 + 2y^2 - 152.
That gradient is <2x, 4y> = k<3,1> for some number k or more compactly 2x = 3(4y) = 12y so x = 6y.
So the points where that is true on the curve satisfy G(x,y)=0 and x=6y.
Now solve G(6y,y) = 0 which is 36y^2 + 2y^2 = 152 or 38y^2 = 152 or 19y^2 = 76 or y^2=4
The solutions are (12,2) and (-12,-2).
It is good because 144+8=152.
The shape is an ellipse so the vector that goes from the origin to (x,y) isn't normal to the tangent as it would be in a circle.
The line 2x - 6y - 7 = 0 has slope 1/3 and a vector along it is <3,1>.
The normals to the curve are parallel to the gradient of the function G(x) = x^2 + 2y^2 - 152.
That gradient is <2x, 4y> = k<3,1> for some number k or more compactly 2x = 3(4y) = 12y so x = 6y.
So the points where that is true on the curve satisfy G(x,y)=0 and x=6y.
Now solve G(6y,y) = 0 which is 36y^2 + 2y^2 = 152 or 38y^2 = 152 or 19y^2 = 76 or y^2=4
The solutions are (12,2) and (-12,-2).
It is good because 144+8=152.
The shape is an ellipse so the vector that goes from the origin to (x,y) isn't normal to the tangent as it would be in a circle.
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2x-6y-7=0
(1/3)x - 7/3 = y and our slope is 1/3, so we have to find normals with this slope.
Since the derivative of a function gives us tangent slopes, we can take the negative multiplicative inverse of the normal slope and find the tangent slope to find. So we have to solve our derivative for -3.
2x + 4y(dy/dx) = 0
dy/dx = -2x/4y = -x/2y
(1/3)x - 7/3 = y and our slope is 1/3, so we have to find normals with this slope.
Since the derivative of a function gives us tangent slopes, we can take the negative multiplicative inverse of the normal slope and find the tangent slope to find. So we have to solve our derivative for -3.
2x + 4y(dy/dx) = 0
dy/dx = -2x/4y = -x/2y
12
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