Question

The locus of the moving point P(x, y) satisfying $\sqrt {{{\left( {x - 1} \right)}^2} + {y^2}} + \sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {y - \sqrt {12} } \right)}^2}} = a$ will be an ellipse if
$(a){\text{ a < 4}} \\\ {\text{(b) a > 2}} \\\ (c){\text{ a > 4}} \\\ {\text{(d) a < 2}} \\\$

Answer 1

Hint: In this question consider an arbitrary point P(x, y) on an ellipse, use distance formula to compare the equation given $\sqrt {{{\left( {x - 1} \right)}^2} + {y^2}} + \sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {y - \sqrt {12} } \right)}^2}} = a$, consider two points $S'{\text{ and S}}$, and find their coordinates. Use the condition of the ellipse that $a > SS'$.

Complete step-by-step solution -

We have,
$\sqrt {{{\left( {x - 1} \right)}^2} + {y^2}} + \sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {y - \sqrt {12} } \right)}^2}} = a$
And point P = (x, y)
Now as we know that the distance between two points ($x_1$, $y_1$) and ($x_2$, $y_2$) is calculated as
$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$
Therefore $\sqrt {{{\left( {x - 1} \right)}^2} + {y^2}}$ is the distance between point P (x, y) and S (1, 0).
Therefore, SP = $\sqrt {{{\left( {x - 1} \right)}^2} + {y^2}}$
Now $\sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {y - \sqrt {12} } \right)}^2}}$ is the distance between point P (x, y) and S’ (-1,$\sqrt {12}$).
Therefore S’P = $\sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {y - \sqrt {12} } \right)}^2}}$
$\Rightarrow SP + S'P = a$
Now the locus of a point P is an ellipse if
$a > SS'$ ............................ (1) Where SS’ is the distance between points (S) and (S’).
As we know S and S’ are the two foci of an ellipse and distance between them is 2Ae, where (e) is the eccentricity and always less than 1 and A is the half length of the major axis of the ellipse.
Therefore, SS’ = 2Ae.................. (2)
So $SP + S'P > 2Ae$ (necessary condition for an ellipse)
From equation (1) and (2) we have,
$\Rightarrow a > SS'$
So, the locus of a point P is an ellipse if $a > SS'$
So the distance between S and S’ is
Let S = ($x_1$, $y_1$) = (1, 0)
And S’ = ($x_2$, $y_2$) = (-1,$\sqrt {12}$)
$\Rightarrow SS' = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} = \sqrt {{{\left( { - 1 - 1} \right)}^2} + {{\left( {\sqrt {12} - 0} \right)}^2}} = \sqrt {4 + 12} = \sqrt {16} = 4$
So from equation (1) we have,
$\Rightarrow a > 4$
So this is the required answer.
Hence option (C) is correct.

Note: Locus is a set of moving points under a certain condition that formulate a curve. For example, a circle is a locus of points that are at a constant distance from a fixed point. This fixed point is termed as the center of the circle and the constant distance is the radius of the circle. In the similar manner it is advised to remember certain locus condition, condition for ellipse is being used above. This helps solve problems of this kind.