Porisma

Worked solution

Recover a rational curve from points and a gradient

The equation of a curve CC is given by y=ax+b+c/xy=ax+b+c/x, where a,b,ca,b,c are constants. The gradient of CC at the point (1,1)(1,1) is 3. It is also given that CC passes through the point (2,2)(2,2). Find the values of a,b,ca,b,c. Hence state the equations of the asymptotes of CC. marks

[2016/YJC/Promo/10(a)]

CJC · Review Problems · Question 6

Answer

a=−1,b=6,c=−4a=-1,b=6,c=-4; asymptotes x=0x=0, y=−x+6y=-x+6.

Translate the three conditions

The points give a+b+c=1a+b+c=1 and 2a+b+12c=22a+b+\tfrac12c=2. Since dy/dx=a−c/x2dy/dx=a-c/x^2, the gradient at x=1x=1 gives a−c=3a-c=3.

How did this part go?

Solve the constants

The source GC solution is a=−1,b=6,c=−4a=-1,b=6,c=-4, so y=−x+6−4/xy=-x+6-4/x. Both original point equations and the derivative condition hold.

How did this part go?

State both asymptotes

The nonzero reciprocal term gives vertical asymptote x=0x=0. As x→±∞x\to\pm\infty, −4/x→0-4/x\to0, so the oblique asymptote is y=−x+6y=-x+6.

How did this part go?

Common mistakes

A gradient condition is an equation in the derivative, not another ordinate value.

Solution notes

Also printed as Review 6 in both CJC Linear Equations lecture copies: question PDF page 14 / printed 174; answer PDF 15 / Student printed 175, Teacher visible footer 15 (extracted 175). Compared in evidence/review-cjc-linear-equations-lecture.md; the repeated asks, tables and marks retain this existing identity.

Read the hint