The minimum area of the triangle formed by th | vedclass

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

Question

(D) The given equation of the line is $3 \cos 	heta \cdot x + 4 \sin 	heta \cdot y = 12$.Dividing by $12$,we get the intercept form:$\frac{x}{4 / \c

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) The given equation of the line is $3 \cos 	heta \cdot x + 4 \sin 	heta \cdot y = 12$.Dividing by $12$,we get the intercept form:$\frac{x}{4 / \cos 	heta} + \frac{y}{3 / \sin 	heta} = 1$.This line intersects the coordinate axes at $A\left(\frac{4}{\cos 	heta}, 0ight)$ and $B\left(0, \frac{3}{\sin 	heta}ight)$.The area of $	riangle OAB$ is given by:$\Delta = \frac{1}{2} 	imes |OA| 	imes |OB| = \frac{1}{2} 	imes \left| \frac{4}{\cos 	heta} ight| 	imes \left| \frac{3}{\sin 	heta} ight| = \frac{6}{|\sin 	heta \cos 	heta|} = \frac{12}{|\sin 2 	heta|}$.For the area to be minimum,$|\sin 2 	heta|$ must be maximum. Since the maximum value of $|\sin 2 	heta|$ is $1$,the minimum area is:$\Delta_{\min} = \frac{12}{1} = 12$.

The minimum area of the triangle formed by the variable line $3 \cos \theta \cdot x + 4 \sin \theta \cdot y = 12$ and the coordinate axes is

Similar Questions

For what values of $a$ and $b$ are the intercepts cut off on the coordinate axes by the line $ax + by + 8 = 0$ equal in length but opposite in sign to those cut off by the line $2x - 3y + 6 = 0$ on the axes?

Reduce the following equation into intercept form and find its intercepts on the axes: $3y + 2 = 0$.

The equation of the line parallel to $x/a + y/b = 1$ and passing through $(a, b)$ is .....

The gradient of the line joining the points on the curve $y = x^2 + 2x$ whose abscissae are $1$ and $3$,is

If the line $y = mx + c$ passes through the points $(2, 4)$ and $(3, -5)$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module