If graph of the expression y=x2-8 x+12 is shown in the figure then area of the square A B C D inscribed between parabola and x-axis is given by

Question

If graph of the expression y=x2-8 x+12 is shown in the figure then area of the square A B C D inscribed between parabola and x-axis is given by (A) 12+4 √5 (B) 12-4 √5 (C) 24+8 √5 (D) 24-8 √5

Tardigrade Admin · Accepted Answer

Let the side length be 2k <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/fc88755d58d9cf5ea7d69a94c79aa7b3-.png" /> Now point ((k+4),-2 k) lies on parabola y=x2-8 x+12. ∴ (-2 k)=(k+4)2-8(k+4)+12 ⇒ k2+2 k-4=0 ⇒ k=(-2+√4+16/2)=-1+√5 So, area =4 k2=4(-1+√5)2=4(6-2 √5)

Q. If graph of the expression $y = x^{2} - 8 x + 12$ is shown in the figure then area of the square $A BC D$ inscribed between parabola and $x$ -axis is given by

$12 + 45$

$12 - 45$

$24 + 85$

$24 - 85$

Let the side length be $2 k$

Now point $((k + 4), - 2 k)$ lies on parabola $y = x^{2} - 8 x + 12$ .
$∴ (- 2 k) = (k + 4)^{2} - 8 (k + 4) + 12$
$\Rightarrow k^{2} + 2 k - 4 = 0$
$\Rightarrow k = \frac{- 2 + 4 + 16}{2} = - 1 + 5$
So, area $= 4 k^{2} = 4 (- 1 + 5)^{2} = 4 (6 - 25)$

Q. If graph of the expression y=x2−8x+12 is shown in the figure then area of the square ABCD inscribed between parabola and x-axis is given by

12+45​

12−45​

24+85​

24−85​