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Question: If from the points (h, 2-5h), h ∈ R, h≠ 1 two distinct tangents are drawn to the curve y = x³-3x² - ...

If from the points (h, 2-5h), h ∈ R, h≠ 1 two distinct tangents are drawn to the curve y = x³-3x² - ax + b then find the value of (a+b).

Answer

3

Explanation

Solution

Solution:

We are given the cubic

f(x)=x3−3x2−ax+b,f(x)=x^3-3x^2-ax+b,

and a family of points

P(h,2−5h)(h∈R, h≠1).P(h,2-5h)\quad (h\in\mathbb{R},\, h\neq 1).

A tangent to f(x)f(x) at any point x=rx=r is

y=f(r)+f′(r)(x−r),withf′(r)=3r2−6r−a.y=f(r)+f'(r)(x-r),\quad\text{with}\quad f'(r)=3r^2-6r-a.

For this tangent to pass through P(h,2−5h)P(h,2-5h) we require

2−5h=f(r)+f′(r)(h−r).2-5h = f(r)+f'(r)(h-r).

Substitute

f(r)=r3−3r2−ar+b,f′(r)=3r2−6r−a.f(r)=r^3-3r^2-ar+b,\quad f'(r)=3r^2-6r-a.

Then

2−5h=r3−3r2−ar+b+(3r2−6r−a)(h−r).2-5h=r^3-3r^2-ar+b+(3r^2-6r-a)(h-r).

Expanding and simplifying (grouping powers of rr) one obtains a cubic in rr whose roots represent points of tangency. In order that exactly two distinct tangents are drawn from P(h,2−5h)P(h,2-5h), the cubic must have a double root and a simple root. Differentiating the cubic (with respect to rr) and eliminating common factors, one finds (after routine algebra) that the double root is either r=hr=h or r=1r=1. The possibility r=hr=h forces a contradiction (since the coefficient of r3r^3 would then be nonzero for arbitrary hh). Hence the double root must be

r=1.r=1.

Thus one of the tangency points is fixed, namely at r=1r=1. We then have

(1, f(1))withf(1)=13−3⋅12−a⋅1+b=1−3−a+b.(1,\, f(1))\quad\text{with}\quad f(1)=1^3-3\cdot1^2-a\cdot 1+b=1-3-a+b.

Since the tangent at this point must pass through P(h,2−5h)P(h,2-5h) for all h≠1h\ne 1, in particular the fixed tangent line at r=1r=1 must contain every point of the form

(h,2−5h).\Big(h,2-5h\Big).

The tangent line at r=1r=1 is found using

f′(1)=3−6−a=−3−a.f'(1)=3-6-a= -3-a.

Its equation is

y−f(1)=f′(1)(x−1).y-f(1)=f'(1)(x-1).

Plug in x=hx=h and y=2−5hy=2-5h:

2−5h−f(1)=(−3−a)(h−1).2-5h - f(1)= (-3-a)(h-1).

Also, at r=1r=1 the point of contact is (1,f(1))(1,f(1)) and substituting x=1x=1 the equation must yield f(1)=f(1)f(1)=f(1). Now, note that for the external point P(h,2−5h)P(h,2-5h) to lie on the tangent, its xx-coordinate hh is arbitrary (with h≠1h\neq 1). This is possible only if the entire family of points lies on the tangent line. But the set

{(h,2−5h):h∈R}\{(h,2-5h):h\in\Bbb R\}

is a straight line with equation

y=−5x+2.y=-5x+2.

Thus the tangent line at r=1r=1 is

y=−5x+2.y=-5x+2.

But we also have from the tangent line at r=1r=1:

y−f(1)=(−3−a)(x−1).y-f(1)=(-3-a)(x-1).

Comparing slopes, we must have

−3−a=−5⟹a=2.-3-a = -5\quad\Longrightarrow\quad a=2.

Now, equate the constant term by noting that the point (1,f(1))(1, f(1)) lies on y=−5x+2y=-5x+2:

f(1)=−5(1)+2=−3.f(1)= -5(1)+2=-3.

But f(1)=1−3−2+b=−4+b.f(1)=1-3-2+b=-4+b. Hence

−4+b=−3⟹b=1.-4+b=-3\quad\Longrightarrow\quad b=1.

Thus,

a+b=2+1=3.a+b=2+1=3.

Final Answer: 3

Summary (Minimal Explanation):

  1. Write the equation of the tangent to f(x)=x3−3x2−ax+bf(x)=x^3-3x^2-ax+b at x=rx=r:

    y=f(r)+f′(r)(x−r)y=f(r)+f'(r)(x-r) with f′(r)=3r2−6r−af'(r)=3r^2-6r-a.

  2. For (h,2−5h)(h,2-5h) to lie on the tangent, set

    2−5h=f(r)+f′(r)(h−r)2-5h=f(r)+f'(r)(h-r).

  3. Requiring that for arbitrary hh (except h=1h=1) there are two distinct tangents forces one of the tangency points to be fixed; it turns out r=1r=1 must be a double root.

  4. Since the tangent at r=1r=1 must contain all (h,2−5h)(h,2-5h), it is identical with the line y=−5x+2y=-5x+2.

  5. Equate the slope −5-5 with −3−a-3-a to get a=2a=2 and use f(1)=−3f(1)=-3 to obtain b=1b=1.

  6. Therefore, a+b=3a+b=3.