First slide

Ellipse

Question

A point moves such that the sum of the square of its distances from two fixed straight lines intersecting at angle 2 $α$ is a constant. The locus of point is an ellipse of eccentricity

Difficult

A

$\frac{\sqrt{\cos 2 α}}{\sin α} if α < \frac{π}{4}$
B

$\frac{\sqrt{- \cos 2 α}}{\cos α} if α > \frac{π}{4}$
C

$\frac{\sqrt{\cos 2 α}}{\cos α} if α < \frac{π}{4}$
D

$\frac{\sqrt{- \cos 2 α}}{\sin α} if α > \frac{π}{4}$

Solution

Let us choose the point of intersection of the given lines as the origin and their angular bisector as the x-axis. Then equation of the two lines will be y = mx and y = -mx, where m = tan $α$ .
Let P(h, k)be the point whose locus is to be found. Then according to the given condition,
PA² + PB² = Constant
$\Rightarrow \frac{(k - mh)^{2}}{1 + m^{2}} + \frac{(k + mh)^{2}}{1 + m^{2}} = c (c is a constant) \Rightarrow 2 (k^{2} + m^{2} h^{2}) = c (1 + m^{2})$
Therefore, the locus of point P is
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, where a^{2} = \frac{c (1 + m^{2})}{2 m^{2}} and b^{2} = \frac{c (1 + m^{2})}{2} . If α < π / 4, then m < 1, and a^{2} > b^{2} . ∴ Eccentricity = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - m^{2}} = \frac{\sqrt{\cos 2 α}}{\cos α} If α > π / 4, then m > 1, and a^{2} < b^{2} . ∴ Eccentricity = \sqrt{1 - \frac{a^{2}}{b^{2}}} = \sqrt{1 - \frac{1}{m^{2}}} = \frac{\sqrt{- \cos 2 α}}{\sin α}$

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