Question

In a square $ABCD$ equation of the $AB$ is $x-y+8=0$ and $C,D$ lies on the curve $y=x^2$

• A. maximum $y$ intercept of $CD=30$
• B. maximum $y$ intercept of $CD=12$
• C. minimum area of $ABCD=18$ square units
• D. maximum area of $ABCD=242$square units

Answer 1

Answer: (A), (C), (D)

The correct options are
A maximum $y$ intercept of $CD=30$
C minimum area of $ABCD=18$ square units
D maximum area of $ABCD=242$square units

Let the coordinates of the point $C(x_1,y_1)\text{and}D(x_2,y_2)$ and $|CD|=a$
Equation of the $CD$ will be
$x-y=b$
$\frac{y_1-y_2}{x_1-x_2}=1$
Also,
$a^2=(x_1-x_2{)}^2+(y_1-y_2{)}^2{a}^2=2\left[\right(x_1+x_2{)}^2-4x_1{x}_2]$
Equating line $CD$ and curve $x^2=y$
$x^2=x-b\Rightarrow x^2-x+b=0$
$x_1+x_2=1x_1{x}_2=b{a}^2=2(1-4b)$
distance of $C$ from the line $AB$
$a=\left|\frac{x_1-y_1+8}{\sqrt{2}}\right|a^2=\frac{(b+8{)}^2}{2}\Rightarrow 4(1-4b)=(b+8{)}^2\Rightarrow b=-2\text{or}-30b=-2\Rightarrow a=3\sqrt{2}b=-30\Rightarrow a=11\sqrt{2}$
maximum area of $ABCD=242uni{t}^2$
minimum area of $ABCD=18uni{t}^2$
maximum $y$ intercept of the line
$CD=30$
minimum $y$ intercept of the line
$CD=2$