Q78 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics
If $(x - a)^2 + (y - b)^2 = c^2$, where a, b, c are some constants, $c > 0$ then $\dfrac{\left[1 + \left(\dfrac{dy}{dx}\right)^2\right]^{\dfrac{3}{2}}}{\dfrac{d^2y}{dx^2}}$ is dependent on:
- Ax
- BConstants a and b
- Cy
- Correct answerDConstant c
Solution
Differentiating $(x-a)^2 + (y-b)^2 = c^2$:
$\dfrac{dy}{dx} = -\dfrac{x-a}{y-b}$
$1 + \left(\dfrac{dy}{dx}\right)^2 = 1 + \dfrac{(x-a)^2}{(y-b)^2} = \dfrac{(x-a)^2+(y-b)^2}{(y-b)^2} = \dfrac{c^2}{(y-b)^2}$
Differentiating again using quotient rule:
$\dfrac{d^2y}{dx^2} = -\dfrac{(y-b)^2 + (x-a)^2}{(y-b)^3} = -\dfrac{c^2}{(y-b)^3}$
Substituting into the expression:
$\dfrac{\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{3/2}}{\dfrac{d^2y}{dx^2}} = \dfrac{\left(\dfrac{c^2}{(y-b)^2}\right)^{3/2}}{-\dfrac{c^2}{(y-b)^3}} = \dfrac{\dfrac{c^3}{(y-b)^3}}{-\dfrac{c^2}{(y-b)^3}} = -c$
The expression depends only on constant $c$.
$\dfrac{dy}{dx} = -\dfrac{x-a}{y-b}$
$1 + \left(\dfrac{dy}{dx}\right)^2 = 1 + \dfrac{(x-a)^2}{(y-b)^2} = \dfrac{(x-a)^2+(y-b)^2}{(y-b)^2} = \dfrac{c^2}{(y-b)^2}$
Differentiating again using quotient rule:
$\dfrac{d^2y}{dx^2} = -\dfrac{(y-b)^2 + (x-a)^2}{(y-b)^3} = -\dfrac{c^2}{(y-b)^3}$
Substituting into the expression:
$\dfrac{\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{3/2}}{\dfrac{d^2y}{dx^2}} = \dfrac{\left(\dfrac{c^2}{(y-b)^2}\right)^{3/2}}{-\dfrac{c^2}{(y-b)^3}} = \dfrac{\dfrac{c^3}{(y-b)^3}}{-\dfrac{c^2}{(y-b)^3}} = -c$
The expression depends only on constant $c$.
Hence, the correct answer is option: 'D'.