Consider the triangles with vertices A(2,1),B | vedclass

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

Question

(A) Let the vertices be $A(2,1)$,$B(0,0)$,and $C(t,4)$ for $t \in [0,4]$. The perimeter $P(t) = AB + BC + AC$. Since $AB = \sqrt{2^2 + 1^2} = \sqrt{5}

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(A) Let the vertices be $A(2,1)$,$B(0,0)$,and $C(t,4)$ for $t \in [0,4]$. The perimeter $P(t) = AB + BC + AC$. Since $AB = \sqrt{2^2 + 1^2} = \sqrt{5}$ is constant,we need to maximize/minimize $f(t) = BC + AC = \sqrt{t^2 + 4^2} + \sqrt{(t-2)^2 + (4-1)^2} = \sqrt{t^2 + 16} + \sqrt{(t-2)^2 + 9}$.To find the minimum,we reflect $B(0,0)$ across the line $y=4$ to get $B'(0,8)$. The minimum occurs when $A, C, B'$ are collinear. The line $AB'$ passes through $(2,1)$ and $(0,8)$,so its equation is $y - 8 = \frac{1-8}{2-0}(x - 0) \implies y = -\frac{7}{2}x + 8$. Setting $y=4$,we get $4 = -\frac{7}{2}t + 8 \implies \frac{7}{2}t = 4 \implies t = \frac{8}{7}$. Thus,$\beta = \frac{8}{7}$.To find the maximum,we examine the endpoints of the interval $t \in [0,4]$. $f(0) = \sqrt{16} + \sqrt{4+9} = 4 + \sqrt{13} \approx 7.6$. $f(4) = \sqrt{16+16} + \sqrt{4+9} = 4\sqrt{2} + \sqrt{13} \approx 5.65 + 3.6 = 9.25$. Since $f(t)$ is a convex function,the maximum occurs at one of the endpoints. Comparing $f(0)$ and $f(4)$,the maximum is at $t=4$. Thus,$\alpha = 4$.Finally,$6\alpha + 21\beta = 6(4) + 21(\frac{8}{7}) = 24 + 3(8) = 24 + 24 = 48$.

Consider the triangles with vertices $A(2,1)$,$B(0,0)$,and $C(t,4)$,where $t \in [0,4]$. If the maximum and the minimum perimeters of such triangles are obtained at $t=\alpha$ and $t=\beta$ respectively,then $6\alpha + 21\beta$ is equal to $.........$.

Similar Questions

$A$ straight line passing through a point $(3, 2)$ cuts the $X$ and $Y$-axes at points $A$ and $B$ respectively. If a point $P(h, k)$ divides $AB$ in the ratio $2: 3$,then the equation of the locus of point $P$ is

If a point $P$ is equidistant from the points $A(a + b, b - a)$ and $B(a - b, a + b)$,find the locus of $P$.

The locus of the centroid of a triangle whose vertices are $(1, 0)$,$(a \cos t, a \sin t)$,and $(b \sin t, -b \cos t)$ is $9x^2 + 9y^2 - 6x = k$. Then,the value of $k$ is equal to

The locus of a point such that the sum of its distances from two given perpendicular lines is equal to $2$ units in the first quadrant is

If $M$ is a point on the line $y=x$ and points $P(0,1), Q(2,0)$ are such that $PM+QM$ is minimum,the coordinates of $M$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module