P(5,2) is a point on the curve y=f(x) and fra | vedclass

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

Question

(C) Given point $P(5,2)$ and slope of the tangent $m_t = \frac{7}{2}$.Equation of the tangent at $P(5,2)$ is $y - 2 = \frac{7}{2}(x - 5) \implies 2y -

Vedclass · Accepted Answer

$\frac{53}{7}$

Vedclass · Answer

(C) Given point $P(5,2)$ and slope of the tangent $m_t = \frac{7}{2}$.Equation of the tangent at $P(5,2)$ is $y - 2 = \frac{7}{2}(x - 5) \implies 2y - 4 = 7x - 35 \implies 7x - 2y = 31$.The $x$-intercept of the tangent is found by setting $y=0$: $7x = 31 \implies x = \frac{31}{7}$. So,the tangent meets the $x$-axis at $A(\frac{31}{7}, 0)$.The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{2}{7}$.Equation of the normal at $P(5,2)$ is $y - 2 = -\frac{2}{7}(x - 5) \implies 7y - 14 = -2x + 10 \implies 2x + 7y = 24$.The $x$-intercept of the normal is found by setting $y=0$: $2x = 24 \implies x = 12$. So,the normal meets the $x$-axis at $B(12, 0)$.The triangle is formed by points $P(5,2)$,$A(\frac{31}{7}, 0)$,and $B(12, 0)$.The base of the triangle on the $x$-axis is $|12 - \frac{31}{7}| = |\frac{84 - 31}{7}| = \frac{53}{7}$.The height of the triangle is the $y$-coordinate of $P$,which is $2$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{53}{7} 	imes 2 = \frac{53}{7}$ sq. units.

$P(5,2)$ is a point on the curve $y=f(x)$ and $\frac{7}{2}$ is the slope of the tangent to the curve at $P$. The area of the triangle (in sq. units) formed by the tangent and the normal to the curve at $P$ with the $x$-axis is:

Similar Questions

The equation of the tangent at $A(2, 3)$ to the curve $y = ax^3 + b$ is $y = 4x - 5$. Then $b =$

If the subnormal at any point on the curve $y^n = ax$ is constant,then the value of $n$ is:

Find points on the curve $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$ at which the tangents are parallel to the $y$-axis.

The coordinates of the point $P$ on the curve $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$,where the tangent is inclined at an angle $\frac{\pi}{4}$ to the $x$-axis,are

If $\theta$ is the angle between the curves $x^2-y^2=4$ and $y^2=3x$,then $\tan \theta=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module