Find equations of all lines tangent to the curve y(x²+4) = 8 when y=1

Question

Please explain the steps in solving this.

Find equations of all lines tangent to the curve y(x²+4) = 8 when y=1

Answer 1

You could use implicit differentiation to find y', but perhaps the easiest path to follow is to solve for y first.

y = 8(x²+4)^-1

Then y' = -8(x²+4)^-2(2x) = -16x/(x²+4)²

When y = 1, x²+4 = 8, x²=4, x = ±2.

When x = 2, y'(2) = -32/64 = -1/2.

When x = -2, y'(-2) = 32/64 = 1/2.

So there are two tangent lines, one passing through (2,1) with a slope of -1/2, and...

a 2nd passing through (-2,1) with a slope of 1/2.

Use the point-slope formula to write the equations of those two lines, perhaps transforming to slope-intercept form.

Note that is you use implicit differentiation instead of solving for y first:

y(2x)+(x²+4)y' = 0

y' = -2xy/(x²+4)

y'(2,1) = -4/8 = -1/2

y'(-2,1) = 4/8 = 1/2.

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