If the equation of the tangent at (2,3) on th | vedclass

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

Question

(B) Given the curve $y^2 = ax^3 + b$. Since the point $(2,3)$ lies on the curve,we have $3^2 = a(2)^3 + b$,which implies $9 = 8a + b$ or $b = 9 - 8a$.

Vedclass · Accepted Answer

$53$

Vedclass · Answer

(B) Given the curve $y^2 = ax^3 + b$. Since the point $(2,3)$ lies on the curve,we have $3^2 = a(2)^3 + b$,which implies $9 = 8a + b$ or $b = 9 - 8a$. Differentiating the equation of the curve with respect to $x$,we get $2y \frac{dy}{dx} = 3ax^2$,so $\frac{dy}{dx} = \frac{3ax^2}{2y}$. At the point $(2,3)$,the slope of the tangent is $\frac{dy}{dx} = \frac{3a(2)^2}{2(3)} = \frac{12a}{6} = 2a$. The equation of the tangent line is given as $y = 4x - 5$,which has a slope of $4$. Equating the slopes,$2a = 4$,so $a = 2$. Substituting $a = 2$ into $b = 9 - 8a$,we get $b = 9 - 8(2) = 9 - 16 = -7$. Finally,$a^2 + b^2 = (2)^2 + (-7)^2 = 4 + 49 = 53$.

If the equation of the tangent at $(2,3)$ on the curve $y^2 = ax^3 + b$ is $y = 4x - 5$,then the value of $a^2 + b^2$ is:

Similar Questions

If the relation between the subnormal $SN$ and the subtangent $ST$ at any point $S$ on the curve $by^2 = (x + a)^3$ is $p(SN) = q(ST)^2$,then what is the value of $p/q$?

The normal to the curve $y=f(x)$ at the point $(3,4)$ makes an angle $\frac{3 \pi}{4}$ with the positive $X$-axis. Then $f^{\prime}(3)$ is equal to:

The equation of the normal to the curve $y = \sin x$ at the point $(0, 0)$ is

If two curves $x=y^2$ and $xy=a^3$ cut each other orthogonally at a point,then $a^2$ is equal to

If the tangent at point $(1, 2)$ on the curve $y = ax^2 + bx + \frac{7}{2}$ is parallel to the normal at $(-2, 2)$ on the curve $y = x^2 + 6x + 10$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module