Finding Points of Tangency for a Given Tangent Line Equation

[BrownianMan]

The function of a line is y^2 + x^3 = 9. I calculated the slope of its tangent to be -3x^2/2y. The question asked us to find a point(s) so that the equation of its tangent line is y + 6x = 13. So it's slope must be -6 at that point.

I got (2,1) as a point. Are there more than one, or is that the only one?

 

[rock.freak667]

well you will have 3x²/2y =6 such that x²=4y

So for each 'x' you pick, there will be a corresponding value of 'y' from that relation. Yes (2,1) is a point, so is (-2,1)

 

[BrownianMan]

But if the point is (-2,1), then the equation for the tangent line can't be y + 6x = 13, because (-2,1) isn't on the line y + 6x = 13.

 

[Mark44]

So here's what you know:
For the curve y^2 + x^3 = 9 (it's not a line, as you stated), y' = -3x^2/(2y).
The slope of the specified tangent line is -6.
Equating those values gives x^2 = 4y ==> x = +/-2sqrt(y)

Now substitute for x in the equation of the tangent line y = -6x + 13. This gives you two values of y, one of which is y = 1. The other is extraneous.

Now substitute this y value in the equation of your curve y^2 + x^3 = 9, and you get only one x-value.

(-2, 1) is not a point on your curve.

 

[BrownianMan]

So the only point is (2,1) as I stated in my first post, right?

 

[Mark44]

Right. (2, 1) and (2, -1) are both points on the graph of the curve, but only the first point has a tangent whose slope is -6.

 

[BrownianMan]

Thank you so much!