Question text 2Marks What is the value of [math: x] where the tangent line to the curve [math: x2+y2+xy=3]x^2+y^2+xy=3 at the point [math: (1,1)] intersects the the [math: x]-axis (i.e. where [math: y=0])? Answer [math: x=] Answer 1[input].Notes Report question issue Question 27 Notes

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BlackTom AI · Accepted Answer

First, I restate the problem: we have the curve x^2 + y^2 + x y = 3, and we want the x-coordinate where the tangent to the curve at the point (1,1) crosses the x-axis (i.e., where y = 0).

To find the slope of the tangent, differentiate implicitly with respect to x. Starting from x^2 + y^2 + x y = 3:
- The derivative of x^2 is 2x.
- The derivative of y^2 is 2y·(dy/dx) = 2y y'.
- The derivative of x y is y + x y' (product rule).
- The derivative of 3 is 0.
Putting it together: 2x + 2y y' + y + x y' = 0.

Group the y' terms and the constants:
(2y + x) y' + (2x + y) = 0.
So the slope y' of the tangent is y' = -(2x + y)/(2y + x).

Evaluate at the given point (x, y) = (1, 1):
y' = -(2·1 + 1)/(2·1 + 1) = -3/3 = -1.

Thus, the tangent line at (1,1) has slope -1. Using the point-slope form, the equation of the tangent is:
y − 1 = −1(x − 1) 
which simplifies to y = −x + 2.

To find where this tangent crosses the x-axis, set y = 0:
0 = −x + 2 ⇒ x = 2.

Therefore, the x-coordinate where the tangent intersects the x-axis is 2. The given answer 3 does not match this result, indicating a possible miscalculation or a mismatch in the supplied answer. Note that all steps follow from implicit differentiation and the standard tangent-line calculation at a point on an implicitly defined curve.

Question text 2Marks What is the value of [math: x] where the tangent line to the curve [math: x2+y2+xy=3]x^2+y^2+xy=3 at the point [math: (1,1)] intersects the the [math: x]-axis (i.e. where [math: y=0])? Answer [math: x=] Answer 1[input].Notes Report question issue Question 27 Notes

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