differentiation graph question

[qwert]

The question goes like this:
A curve has equation

Find the equations of the normal to the curve which are parallel to the line

The answer is 3y = -x -16√ 3
and 3y= -x+16√ 3


Thanks! And if possible tell me whats wrong with my working?
This might be a little long ><..

so gradient of the normal of the curve would be -1/3 (which is the dy/dx of the line its parallel to)
Since the normal's gradient is -1/3, the tangent's gradient would then be 3.
By knowing this, i equated 3 to the dy/dx of the curve.
And by which i then got ,
3 = 1/4 (3x^2 -24)
So, x = √ 12 or -√ 12

I sub these into the equation of the curve,
which would get me :

y= 2√ 3 - 12√ 3
= -10 √ 3

And then i got stuck....


Thank you so much! I'll appreciate it a lot!

 

[Deleted member 4993]

The question goes like this:
A curve has equation

Find the equations of the normal to the curve which are parallel to the line

The answer is 3y = -x -16√ 3
and 3y= -x+16√ 3


Thanks! And if possible tell me whats wrong with my working?
This might be a little long ><..

so gradient of the normal of the curve would be -1/3 (which is the dy/dx of the line its parallel to)
Since the normal's gradient is -1/3, the tangent's gradient would then be 3.
By knowing this, i equated 3 to the dy/dx of the curve.
And by which i then got ,
3 = 1/4 (3x^2 -24)
So, x = √ 12 or -√ 12

I sub these into the equation of the curve,
which would get me :

y= 2√ 3 - 12√ 3
= -10 √ 3

And then i got stuck....


Thank you so much! I'll appreciate it a lot!

Good work...

So now you know that the tangents pass through (√12, -10√3) or (-√12, +10√3) and slope of 3.

What is the equation of the line that passes through (x₁, y₁) and has a slope of 'm'? → y - y₁ = m*(x - x₁)

 

[srmichael]

The question goes like this:
A curve has equation

Find the equations of the normal to the curve which are parallel to the line

The answer is 3y = -x -16√ 3
and 3y= -x+16√ 3


Thanks! And if possible tell me whats wrong with my working?
This might be a little long ><..

so gradient of the normal of the curve would be -1/3 (which is the dy/dx of the line its parallel to)
Since the normal's gradient is -1/3, the tangent's gradient would then be 3.
By knowing this, i equated 3 to the dy/dx of the curve.
And by which i then got ,
3 = 1/4 (3x^2 -24)
So, x = √ 12 or -√ 12

I sub these into the equation of the curve,
which would get me :

y= 2√ 3 - 12√ 3
= -10 √ 3

And then i got stuck....


Thank you so much! I'll appreciate it a lot!

Careful! Check your math. The two points you should get are \(\displaystyle (2\sqrt{3},-6\sqrt{3})\ and\ (-2\sqrt{3},6\sqrt{3})\)