If the tangent to the curve xy+ax+by=0 at (1, | vedclass

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

Question

(B) The given curve is $xy+ax+by=0$. Since the curve passes through $(1,1)$,we have $1(1)+a(1)+b(1)=0$,which implies $a+b=-1$ (Equation $1$). Differen

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) The given curve is $xy+ax+by=0$. Since the curve passes through $(1,1)$,we have $1(1)+a(1)+b(1)=0$,which implies $a+b=-1$ (Equation $1$). Differentiating the equation of the curve with respect to $x$,we get $x \frac{dy}{dx} + y + a + b \frac{dy}{dx} = 0$. Rearranging for $\frac{dy}{dx}$,we get $\frac{dy}{dx}(x+b) = -(y+a)$,so $\frac{dy}{dx} = -\frac{y+a}{x+b}$. At the point $(1,1)$,the slope of the tangent is $m = \left(\frac{dy}{dx}ight)_{(1,1)} = -\frac{1+a}{1+b}$. Given that the tangent makes an angle $	an^{-1} 2$ with the $x$-axis,the slope $m = 	an(	an^{-1} 2) = 2$. Equating the two expressions for the slope: $-\frac{1+a}{1+b} = 2$,which simplifies to $-(1+a) = 2(1+b)$,or $a+2b = -3$ (Equation $2$). Subtracting Equation $1$ from Equation $2$: $(a+2b) - (a+b) = -3 - (-1)$,which gives $b = -2$. Substituting $b = -2$ into Equation $1$: $a - 2 = -1$,so $a = 1$. Finally,$\frac{a+b}{ab} = \frac{1+(-2)}{(1)(-2)} = \frac{-1}{-2} = \frac{1}{2}$.

If the tangent to the curve $xy+ax+by=0$ at $(1,1)$ makes an angle $\tan^{-1} 2$ with the $x$-axis,then $\frac{a+b}{ab} =$

Similar Questions

Find the slope of the tangent to the curve $y=3x^{4}-4x$ at $x=4$.

At which point does the line $\frac{x}{a} + \frac{y}{b} = 1$ touch the curve $y = be^{-x/a}$?

The point$(s)$ on the curve $y^3 + 3x^2 = 12y$ where the tangent is vertical (parallel to $y$-axis) is (are):

At which points on the curve $y = x^3 - 3x^2 - 9x + 7$ is the tangent parallel to the $x$-axis?

The curve $y=ax^3+bx^2+cx+5$ touches the $X$-axis at $P(-2,0)$ and cuts the $Y$-axis at a point $Q$,where the gradient is $3$. Then the values of $a, b, c$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module