at what points of the ellipse 16x^2 + 9y^2 = 400 does the ordinate decrease at the same rate at - Maths - Application of Derivatives

Question

at what points of the ellipse 16x^2 + 9y^2 = 400 does the
ordinate decrease at the same rate at which the abcissa increases
??

Vimala · Accepted Answer

Let (x, y) be the required point Then $\frac{ⅆ y}{ⅆ t} = - \frac{ⅆ x}{ⅆ t}$ Consider the given curve $16 x^{2} + 9 y^{2} = 400$ (!) Differentiate the above function with respect to x, $32 x \cdot \frac{ⅆ x}{ⅆ t} + 18 y \cdot \frac{ⅆ y}{ⅆ t} = 0$ Substitute the value $\frac{ⅆ y}{ⅆ t} = - \frac{ⅆ x}{ⅆ t}$ in the above equation, we have $32 x \cdot \frac{ⅆ x}{ⅆ t} - 18 y \cdot \frac{ⅆ x}{ⅆ t} = 0 (32 x - 18 y) \cdot \frac{ⅆ x}{ⅆ t} = 0 32 x - 18 y = 0 \Rightarrow y = \frac{16}{9} x$ Now substitute, $y = \frac{16}{9} x$ in the equation of the curve (1), we have $16 x^{2} + 9 {(\frac{16 x}{9})}^{2} = 400 x^{2} = 9 x = \pm 3$ Now for the value $x = 3$ , $y = \frac{16}{9} \times 3 = \frac{16}{3}$ Now for the value $x = - 3$ , $y = \frac{16}{9} \times - 3 = - \frac{16}{3}$ Hence the required points are $(3, \frac{16}{3}) a n d (- 3, \frac{- 16}{3})$

at what points of the ellipse 16x^2 + 9y^2 = 400 does the ordinate decrease at the same rate at which the abcissa increases ??