Question

The minimum distance between the circle ${x^2} + {y^2} = 9$ and the curve $2{x^2} + 10{y^2} + 6xy = 1$ is ?
A. $2\sqrt 2$
B. 2
C. $3 - \sqrt 2$
D. $3 - \dfrac{1}{{\sqrt {11} }}$

Answer 1

We will begin by proving that the normal of the curve $2{x^2} + 10{y^2} + 6xy = 1$ passes through the origin. Then, we will find the maximum distance from the origin to the curve. Then, subtract the distance from origin to curve from the radius to get the required answer.

Complete step by step solution:
On comparing the equation with the standard equation of the circle ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$, where $\left( {h,k} \right)$ are the coordinates of centre and $r$ is the radius of the circle.
Therefore, the coordinates of the centre of ${x^2} + {y^2} = 9$ are $\left( {0,0} \right)$ and the radius is 3 units.
We have to find the centre of the curve.
Let the equation of curve be $g\left( {x,y} \right) = 2{x^2} + 10{y^2} + 6xy = 1$ ……eqn. (1)
Therefore, we will find the normal to the curve which is given by $\nabla g\left( {x,y} \right) = 0$
First calculate the partial differentiation with respect to $x$ and equate it to 0
${g_x} = 4x + 6y$
$\Rightarrow 4x + 6y = 0$ ……eqn. (2)
Similarly, calculate the partial differentiation with respect to $y$ and equate it to 0
${g_y} = 20y + 6x$
$\Rightarrow 20y + 6x = 0$ ……eqn. (3)
The centre of the curve will be the intersection of the normal.
On solving equation (2) and (3) we will get, $\left( {x,y} \right) = \left( {0,0} \right)$
Hence, the centre of the curve is also the origin.
The distance between the circle and the curve is radius of the circle minus the distance from the origin to the curve.
The distance from a point $\left( {x,y} \right)$ on the curve to the origin is
$d = \sqrt {{x^2} + {y^2}} \\\ {d^2} = {x^2} + {y^2} \\\$
Such that $d$ is maximum when ${d^2}$ is maximum.
We will find the maximum value.
$\nabla f\left( {x,y} \right) = \lambda \nabla g\left( {x,y} \right)$
Then,
${f_x} = \lambda {g_x}$
${f_x} = 2x$ and ${g_x} = 4x + 6y$
Hence,
$\Rightarrow 2x = \lambda \left( {4x + 6y} \right) \\\ \Rightarrow \lambda = \dfrac{x}{{2x + 3y}} \\\$
Therefore, ${f_y} = \lambda {g_y}$
${f_y} = 2y$ and ${g_y} = 20y + 6x$
Hence,
$\Rightarrow 2y = \lambda \left( {20y + 6x} \right) \\\ \Rightarrow \lambda = \dfrac{x}{{10y + 3x}} \\\$
On comparing the value of $\lambda$, we will get,
$\dfrac{x}{{2x + 3y}} = \dfrac{x}{{10y + 3x}}$
Cross multiply and simplify the expression,
$\left( {10y + 3x} \right)x = \left( {2x + 3y} \right)y \\\ \Rightarrow 10xy + 3{x^2} = 2xy + 3{y^2} \\\ \Rightarrow 3{x^2} + 8xy - 3{y^2} = 0 \\\$
On completing the square we will get,
${\left( {\sqrt 3 x} \right)^2} + 2\left( {\sqrt 3 x} \right)\left( {\dfrac{4}{{\sqrt 3 }}y} \right) + {\left( {\dfrac{4}{{\sqrt 3 }}y} \right)^2} - 3{y^2} = 0 \\\ \Rightarrow {\left( {\sqrt 3 x + \dfrac{4}{{\sqrt 3 }}y} \right)^2} = \dfrac{{25}}{3}{y^2} \\\$
Take square root and find the value of $x$ and $y$
$\sqrt 3 x + \dfrac{4}{{\sqrt 3 }}y = \pm \dfrac{5}{{\sqrt 3 }}y \\\ \Rightarrow \sqrt 3 x = - \dfrac{4}{{\sqrt 3 }}y \pm \dfrac{5}{{\sqrt 3 }}y \\\ \Rightarrow x = \left( { - 4 \pm 5} \right)\dfrac{y}{3} \\\ \Rightarrow x = \dfrac{1}{3}y, - 3y \\\$
Put the above values in the equation $g\left( {x,y} \right) = 0$
When $x = \dfrac{1}{3}$
${x^2} = \dfrac{{{y^2}}}{9}$
$\Rightarrow 2\dfrac{{{y^2}}}{9} + 10{y^2} + 6y\dfrac{y}{3} - 1 = 0 \\\ \Rightarrow 2{y^2} + 90{y^2} + 18{y^2} - 9 = 0 \\\ \Rightarrow {y^2} = \dfrac{9}{{110}} \\\$
Then, the value of $y = \pm \dfrac{3}{{\sqrt {110} }}$, and $x = \pm \dfrac{1}{{\sqrt {110} }}$
And $x = 3y$
Then, ${x^2} = 9{y^2}$
$\Rightarrow 18{y^2} + 10{y^2} - 18y = 1 \\\ \Rightarrow {y^2} = \dfrac{1}{{10}} \\\$
$y = \pm \dfrac{1}{{\sqrt {10} }}$
Then, $x = \mp \dfrac{3}{{\sqrt {10} }}$
The critical points are $\left( { \pm \dfrac{1}{{\sqrt {110} }}, \pm \dfrac{3}{{\sqrt {110} }}} \right)$ and $\left( { \pm \dfrac{3}{{\sqrt {10} }}, \mp \dfrac{1}{{\sqrt {10} }}} \right)$
From the first set of points, ${d^2} = {x^2} + {y^2} = \dfrac{1}{{110}} + \dfrac{9}{{110}} = \dfrac{1}{{11}}$
And from the set of second point is ${d^2} = {x^2} + {y^2} = \dfrac{1}{{10}} + \dfrac{9}{{10}} = 1$
The maximum value is 1
Hence, the minimum distance is $3 - 1 = 2$

Hence, option B is correct.

Note:
The square of distance as the function is maximised and the point must lie on the curve. The students must know how to take partial derivatives. Concept of Lagrange’s multipliers is used. Also, calculation should be done correctly.