How do you use implicit differentiation to find the points on the curve #x^2+y^2=5x+4y# where tangent line is horizontal and where tangent line is vertical?

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Answer 1 · 2017-07-07T15:17:50

Horizontal tagts. at #(5/2,(4+sqrt41)/2), &, (5/2,(4-sqrt41)/2).#

Eqn. of the Curve C is given to be, # C :x^2+y^2-5x-4y=0.#

Diff.ing w.r.t. #x,# using the Chain Rule, we get,

# 2x+2ydy/dx-5-4dy/dx=0.#

# :. dy/dx=(5-2x)/{2(y-2)}.#

We know that, #dy/dx# denotes the Slope of a tangent to the

Curve at the pt.#(x,y).#

Now, the slope of Horizontal Tangent is #0.#

#:.," for horizontal tgt., "dy/dx=0 rArr (5-2x)/{2(y-2)}=0.#

# :. x=5/2," &, for the corresponding y, we appeal to "C," to get,"#

# x=5/2, C : 25/4+y^2-25/2-4y=0, i.e., y^2-4y-25/4=0.#

#:. y={4+-sqrt(16+25)}/2=(4+-sqrt41)/2.#

Thus, there are two horizontal tagts. at

#(5/2,(4+sqrt41)/2), &, (5/2,(4-sqrt41)/2).#

Similarly, for the Vertical Tangent, we must have undefined

#dy/dx, i.e., y=2,# &, the corresponding #x# can be obtained from #C.#