If the slope of a tangent to the curve xy + a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the curve is $xy + ax + by = 0$. Since the point $(1, 1)$ lies on the curve,we substitute $x = 1$ and $y = 1$: $1(1) + a(1) + b(1) = 0 \impl

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) Given the curve is $xy + ax + by = 0$. Since the point $(1, 1)$ lies on the curve,we substitute $x = 1$ and $y = 1$: $1(1) + a(1) + b(1) = 0 \implies a + b = -1$ ... $(i)$ Now,differentiate the equation $xy + ax + by = 0$ with respect to $x$: $x \frac{dy}{dx} + y + a + b \frac{dy}{dx} = 0$ $(x + b) \frac{dy}{dx} = -(y + a)$ $\frac{dy}{dx} = -\frac{y + a}{x + b}$ Given that the slope at $(1, 1)$ is $2$: $2 = -\frac{1 + a}{1 + b}$ $2(1 + b) = -(1 + a)$ $2 + 2b = -1 - a$ $a + 2b = -3$ ... $(ii)$ Subtracting equation $(i)$ from equation $(ii)$: $(a + 2b) - (a + b) = -3 - (-1)$ $b = -2$ Substituting $b = -2$ into equation $(i)$: $a - 2 = -1 \implies a = 1$ Therefore,$a - b = 1 - (-2) = 3$.

If the slope of a tangent to the curve $xy + ax + by = 0$ at the point $(1, 1)$ on it is $2$,then $a - b$ is

Similar Questions

If $2^x + 2^y = 2^{x + y}$,then $\frac{dy}{dx}$ has the value equal to:

If $e^y + xy = e$,the ordered pair $\left(\frac{dy}{dx}, \frac{d^2y}{dx^2}\right)$ at $x = 0$ is equal to

If $y$ is a function of $x$ and $\log (x+y)=2xy$,then the value of $y^{\prime}(0)$ is

On differentiation, if we obtain $f(x, y) dy - g(x, y) dx = 0$ from $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$, then find the value of $\frac{g(2, 2)}{f(1, 1)}$.

If $x^{x}=y^{y}$,then $\frac{d y}{d x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module