Question

Let $\vec{r}$ be a position vector of a variable point in Cartesian $OXY$ plane such that $\vec{r}\cdot (10\hat{j}-8\hat{i}-\hat{r})=40$ and $p_1=\max \left\{|\vec{r}+2\hat{i}-3\hat{j}{|}^2\right\},p_2=\min \left\{|\vec{r}+2\hat{i}-3\hat{j}{|}^2\right\}$ A tangent line is drawn to the curvy $y=8/x^2$ at point $A$ with abscissa $2$ . The drawn line cuts the $x$-axis at a point $B$.
$p_2$ is equal to

• A. $9$
• B. $2\sqrt{2}-1$
• C. $6\sqrt{2}+3$
• D. $9-4\sqrt{2}$

Answer 1

Answer: (D)

Let $\vec{r}=x\hat{i}+\hat{y}\hat{j}$
$x^2+y^2+8x-10y+40=0$, Which is a circel centre
C $(-4,5)$, radius $r=1$
$p_1=\max \left\{(x+2{)}^2+(y-3{)}^2\right\}$
$p_2=\min \left\{(x+2{)}^2+(y-3{)}^2\right\}$
Let $P$ be $(-2,3)$. Then
$CP=\sqrt{2},r=1$
$p_2=(2\sqrt{2}-1{)}^2$
$p_1=(2\sqrt{2}+1{)}^2$
$p_1+p_2=18$
Slope $=AB=\left(\frac{dy}{dx}\right)=-2$
Equation of $AB.0x+y=6$
$\overrightarrow{OA}=2\hat{i}+2\hat{j},\overrightarrow{OB}=3\hat{i}$
$\overrightarrow{AB}=\hat{i}-2\hat{j}$
$\overrightarrow{AB}\cdot \overrightarrow{OB}=(\hat{i}-2\hat{j})(3\hat{i})=3$