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 $\vec{ r }= x \hat{ i }+\hat{ y } \hat{ j }$
$x^{2}+y^{2}+8 x-10 y+40=0$, 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 $=A B=\left(\frac{d y}{d x}\right)=-2$
Equation of $A B .0 x+y=6$
$\vec{ OA }=2 \hat{ i }+2 \hat{ j }, \vec{ OB }=3 \hat{ i }$
$\vec{ AB }=\hat{ i }-2 \hat{ j } $
$\vec{ AB } \cdot \vec{ OB }=(\hat{ i }-2 \hat{ j })(3 \hat{ i })=3$