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

Find the $x$-coordinate of the point on the curve $ay^2 = x^3$ where the normal makes equal intercepts on the axes.

The equation of the tangent to the curve $y = \sqrt{2} \sin \left(2x + \frac{\pi}{4}\right)$ at $x = \frac{\pi}{4}$ is:

The equation of the tangent to the curve $y = \cos(x + y)$ for $-2\pi \leq x \leq 2\pi$,which is parallel to the line $x + 2y = 0$,is:

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

Find the slope of the tangent to the curve $y = x^{3} - 3x + 2$ at the point whose $x$-coordinate is $3$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module