If the line y = 4x - 5 is tangent to the curv | vedclass

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

Question

(D) Step $1$: Since the point $(2, 3)$ lies on the curve $y^2 = ax^3 + b$, we have $3^2 = a(2)^3 + b$, which simplifies to $9 = 8a + b$ (Equation $1$)

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(D) Step $1$: Since the point $(2, 3)$ lies on the curve $y^2 = ax^3 + b$, we have $3^2 = a(2)^3 + b$, which simplifies to $9 = 8a + b$ (Equation $1$).Step $2$: Differentiate the curve equation with respect to $x$: $2y \frac{dy}{dx} = 3ax^2$, so $\frac{dy}{dx} = \frac{3ax^2}{2y}$.Step $3$: The slope of the tangent at $(2, 3)$ is $\frac{3a(2)^2}{2(3)} = \frac{12a}{6} = 2a$. Given the line $y = 4x - 5$ has a slope of $4$, we equate $2a = 4$, giving $a = 2$.Step $4$: Substitute $a = 2$ into Equation $1$: $9 = 8(2) + b \implies 9 = 16 + b \implies b = -7$.Step $5$: Calculate $7a - 2b = 7(2) - 2(-7) = 14 + 14 = 28$.

If the line $y = 4x - 5$ is tangent to the curve $y^2 = ax^3 + b$ at the point $(2, 3)$, then the value of $7a - 2b$ is...

Similar Questions

The points on the curve $y^2 = x + \sin x$ at which the normal is parallel to the $Y$-axis lie on

What is the angle between the tangents to the curve $y = x^2 - 5x + 6$ at the points $(2, 0)$ and $(3, 0)$?

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 $y=\pi e^{\frac{-x}{\pi}}$ at the point where it crosses the $y$-axis is

If the length of the subnormal at any point on the curve $y^n = a^{n-1}x$ is constant,then what is the value of $n$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module