TMUA practice: Straight Lines, problem 5

A TMUA-calibre straight lines problem at difficulty 4 of 5, with the full worked solution.

The question

A straight line segment

AB

of length

a

moves with end

A

on the

y

-axis and end

B

on the

x

-axis. The locus of the point

P

on the segment for which

AP : BP = 2 : 1

9(x^2 + y^2) = 4a^2

9(x^2 + 4y^2) = 4a^2

9(y^2 + 4x^2) = 4a^2

9x^2 + 4y^2 = a^2

The correct answer is highlighted. B

Worked solution

Let $A = (0, \alpha)$ on the $y$ -axis and $B = (\beta, 0)$ on the $x$ -axis. The segment has length $a$ , so

$\alpha^2 + \beta^2 = a^2.$

Locating $P$ . $P$ lies on $AB$ with $AP : PB = 2 : 1$ , so $P$ divides $AB$ two-thirds of the way from $A$ to $B$ :

P = A + \tfrac23(B - A) = (0, \alpha) + \tfrac23(\beta, -\alpha) = \left(\tfrac{2\beta}{3},\ \tfrac{\alpha}{3}\right).

Eliminate $\alpha, \beta$ . Writing $x = \dfrac{2\beta}{3}$ and $y = \dfrac{\alpha}{3}$ gives $\beta = \dfrac{3x}{2}$ and $\alpha = 3y$ . Substituting into $\alpha^2 + \beta^2 = a^2$ :

$9y^2 + \frac{9x^2}{4} = a^2.$

Multiplying by $4$ : $\;36y^2 + 9x^2 = 4a^2$ , that is $9(x^2 + 4y^2) = 4a^2$ .

Answer: B.