The line x + y = 0 touches the curve y^2 = ax | vedclass

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

Question

(C) Step $1$: Since the point $(1, -1)$ lies on the curve $y^2 = ax^3 + b$, we have $(-1)^2 = a(1)^3 + b$, which simplifies to $a + b = 1$. Step $2$:

Vedclass · Accepted Answer

$a = 2/3, b = 1/3$

Vedclass · Answer

(C) Step $1$: Since the point $(1, -1)$ lies on the curve $y^2 = ax^3 + b$, we have $(-1)^2 = a(1)^3 + b$, which simplifies to $a + b = 1$. Step $2$: Differentiate the curve equation with respect to $x$: $2y \frac{dy}{dx} = 3ax^2$. Step $3$: At the point $(1, -1)$, the slope of the tangent is $\frac{dy}{dx} = \frac{3a(1)^2}{2(-1)} = -\frac{3a}{2}$. Step $4$: The line $x + y = 0$ can be written as $y = -x$, which has a slope of $-1$. Step $5$: Equating the slopes: $-\frac{3a}{2} = -1$, which gives $a = 2/3$. Step $6$: Substitute $a = 2/3$ into $a + b = 1$: $2/3 + b = 1$, so $b = 1/3$.

The line $x + y = 0$ touches the curve $y^2 = ax^3 + b$ at the point $(1, -1)$. Find the values of $a$ and $b$.

Similar Questions

The normals to the curve $y=x^{2}-x+1$, drawn at the points with the abscissae $x_{1}=0, x_{2}=-1$ and $x_{3}=5/2$, are:

If the tangent at the point $(x_{1}, y_{1})$ on the curve $y=x^{3}+3x^{2}+5$ passes through the origin,then $(x_{1}, y_{1})$ does $NOT$ lie on which of the following curves?

Find the equation of the tangent to the curve $y + \frac{2}{x - 3} = 0$ which has a slope of $2$.

If the curves $y^2 = 12x - 3$ and $y^2 = 12 - kx$ cut each other orthogonally, then the length of the sub-tangent at $(1, b)$ on the curve $y^2 = 12 - kx$ is

The tangent drawn at the point $(0, 1)$ on the curve $y = e^{2x}$ meets the $x-$axis at the point:

Vedclass Test Series

Exam Paper Generator

Online Exam Module