First slide

Cartesian plane

Question

$The line joining (5, 0) to (10 \cos θ, 10 \sin θ) is divided internally in the ratio 2 : 3 at P . The locus of P is$

Moderate

A

$x^{2} + 2 x y + y^{2} - 6 x = 0$
B

$x + y - 3 = 0$
C

$(x - 3)^{2} + y^{2} = 16$
D

$x^{2} = y - 3$

Solution

$P (h, k) divides A (5, 0) and B (10 \cos θ, 10 \sin θ) internally in$ $the ratio 2 : 3 .$
$∴ (\frac{15 + 20 \cos θ}{5}, \frac{20 \sin θ}{5}) = (h, k)$
$∴ 15 + 20 \cos θ = 5 h and 20 \sin θ = 5 k$
$∴ 20 \cos θ = 5 h - 15 and 20 \sin θ = 5 k$
Squaring and adding, we get
$\begin{aligned} 400 = (5 h - 15)^{2} + (5 k)^{2} \\ ∴ Locus is (x - 3)^{2} + y^{2} = 16 \end{aligned}$

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