If the tangent at point (1, 2) on the curve y | vedclass

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

Question

(A) For the curve $y = x^2 + 6x + 10$,the derivative is $\frac{dy}{dx} = 2x + 6$.At point $(-2, 2)$,the slope of the tangent is $m_t = 2(-2) + 6 = 2$.

Vedclass · Accepted Answer

$a=1$

Vedclass · Answer

(A) For the curve $y = x^2 + 6x + 10$,the derivative is $\frac{dy}{dx} = 2x + 6$.At point $(-2, 2)$,the slope of the tangent is $m_t = 2(-2) + 6 = 2$.Therefore,the slope of the normal at $(-2, 2)$ is $m_n = -\frac{1}{m_t} = -\frac{1}{2}$.For the curve $y = ax^2 + bx + \frac{7}{2}$,the point $(1, 2)$ lies on the curve,so $2 = a(1)^2 + b(1) + \frac{7}{2}$,which simplifies to $a + b = 2 - \frac{7}{2} = -\frac{3}{2}$ (Equation $1$).The derivative is $\frac{dy}{dx} = 2ax + b$. At $x = 1$,the slope of the tangent is $2a + b$.Since this tangent is parallel to the normal of the first curve,$2a + b = -\frac{1}{2}$ (Equation $2$).Subtracting Equation $1$ from Equation $2$: $(2a + b) - (a + b) = -\frac{1}{2} - (-\frac{3}{2}) \Rightarrow a = 1$.Substituting $a = 1$ into Equation $1$: $1 + b = -\frac{3}{2} \Rightarrow b = -\frac{3}{2} - 1 = -\frac{5}{2}$.

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:

Similar Questions

Find the coordinates of the point on the curve $\sqrt{x}+\sqrt{y}=4$ at which the tangent is equally inclined to the axes.

$V$ is the set of points on the curve $y^3 - 3xy + 2 = 0$ where the tangent is vertical. Then $V = $

The angle at which the curve $y = K e^{Kx}$ intersects the $y$-axis is

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

If $y=2x$ is a tangent to the curve $y^2=ax^3+b$ at $(1,2)$, then $(a, b)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module