Question

Solve the following : Find all points on the ellipse 9x² + 16y² = 400, at which the y-coordinate is decreasing and the coordinate is increasing at the same rate.

Answer 1

L P(x₁, y₁) be the point on the ellipse 9x² + 16y² = 400, at which the y-coordinate decreasing and the x-coordinate is increasing at the same rate..

Then `-(dy/dt)_("at" (x_1, y_1)) = (dx/dy)_("at" (x_1, y_1)`    ...(1)

Differentiating 9x² + 16y² = 400 w.r.t. t, we get

`9 xx 2xdx/dt + 16 xx 2ydy/dx` = 0

∴ `9xdx/dt + 16ydy/dt` = 0

∴ `9x_1(dx/dt)_("at" (x_1, y_1)) + 16y_1(dy/dt)_("at"(x_1, y_1)` = 0

∴ `9x_1(dx/dt)_("at" (x_1, y_1)) - 16y_1(dy/dt)_("at"(x_1, y_1)` = 0     ..[By (1)]

∴ 9x1 –116y1 = 0                            ...(2)
Now, (x1, y1) lies on the ellipse 9x² + 16y² = 400

∴ 9x₁² + 16y₁² = 400

From (2), `x_1 =(16y_1)/(9)`

Substitute `x_1 = (16y_1)/(9)` in (3), we get

∴ `9((16y_1)/9)^2 + 16_1^2` = 400

∴ 16y₁² + 9y₁² = 225

∴ 25y₁² = 225

∴ y₁² = 9

∴ y₁ = ± 3
When y₁ = 3, `x_1 = (16(3))/(9) = (16)/(3)`

When y₁ = – 3, `x_1 = (16(-3))/(9) = -(16)/(3)`
Hence, the required points on the ellipse are

`(16/3, 3) and (-16/3, -3)`.