1.

The points of the ellipse `16x^2+9y^2=400` at which the ordinate decreases at the same rate at which the abcissa increases isA. `(3, 16//3)`B. `(-3, 16//3)`C. `(3, 16//3)`D. `(3, -3)`

Answer» Correct Answer - A
We have,
`16x^(2)+9y^(2)=400`
`implies 32x(dx)/(dt)+18 y (dy)/(dt)=0 " [Differentiating w.r.t. t]"`
`implies 32x(dx)/(dt)-18y(dx)/(dt)=0 " "[because-(dy)/(dt)=(dx)/(dt)("given")]`
`implies 16x-9y=0`
`impliesy=(16x)/(9)`
`therefore 16x^(2)+9y^(2)=400 implies 16x^(2)+(256)/(9)x^(2)=400impliesx=pm3`
Now, `y=(16)/(9)x and x=pm3impliesy= pm(16)/(3)`
Hence, the required points are `(3, 16//3)` and `(-3, -16//3)`


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