Question

If the tangent at a point 'P' on an ellipse meets a directrix at Q(3,4) and the equation of line through P and S, S being corresponding focus is $x+2y-11+2\sqrt{5}=0$, such that PSQR forms a square, then area of PSQR is _____.

Answer 1

4

Answer 2

Let P be a point on the ellipse, S be the corresponding focus, and Q be the point where the tangent at P meets the directrix. We are given that PSQR forms a square. This means the vertices are in order, so the sides are PS, SQ, QR, RP. In a square PSQR, the angle $\angle SPQ$ must be $90^\circ$, and the side lengths must be equal, i.e., $PS = PQ$.

We are given the equation of the line PS as $x+2y-11+2\sqrt{5}=0$. The slope of the line PS is $m_{PS} = -\frac{1}{2}$. Since $\angle SPQ = 90^\circ$, the line PQ is perpendicular to the line PS. The slope of the line PQ is $m_{PQ} = -\frac{1}{m_{PS}} = -\frac{1}{-1/2} = 2$.

The line PQ is the tangent at P and passes through Q(3,4). The equation of the line PQ is $y - 4 = 2(x - 3)$. $y - 4 = 2x - 6$ $2x - y - 2 = 0$.

The point P is the intersection of the lines PS and PQ. We solve the system of equations:

1. $x+2y-11+2\sqrt{5}=0$
2. $2x-y-2=0$ From equation (2), $y = 2x - 2$. Substitute this into equation (1): $x + 2(2x - 2) - 11 + 2\sqrt{5} = 0$ $x + 4x - 4 - 11 + 2\sqrt{5} = 0$ $5x - 15 + 2\sqrt{5} = 0$ $5x = 15 - 2\sqrt{5}$ $x_P = \frac{15 - 2\sqrt{5}}{5} = 3 - \frac{2\sqrt{5}}{5}$. Now find $y_P$: $y_P = 2x_P - 2 = 2(3 - \frac{2\sqrt{5}}{5}) - 2 = 6 - \frac{4\sqrt{5}}{5} - 2 = 4 - \frac{4\sqrt{5}}{5}$. So, the coordinates of P are $(3 - \frac{2\sqrt{5}}{5}, 4 - \frac{4\sqrt{5}}{5})$.

Q is given as (3, 4). The length of the side PQ of the square is the distance between P and Q. $PQ = \sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2}$ $x_Q - x_P = 3 - (3 - \frac{2\sqrt{5}}{5}) = \frac{2\sqrt{5}}{5}$ $y_Q - y_P = 4 - (4 - \frac{4\sqrt{5}}{5}) = \frac{4\sqrt{5}}{5}$ $PQ^2 = (\frac{2\sqrt{5}}{5})^2 + (\frac{4\sqrt{5}}{5})^2 = \frac{4 \cdot 5}{25} + \frac{16 \cdot 5}{25} = \frac{20}{25} + \frac{80}{25} = \frac{100}{25} = 4$. $PQ = \sqrt{4} = 2$.

Since PSQR is a square with side length $s = PQ$, the side length is 2. The area of the square PSQR is $s^2 = 2^2 = 4$.

Note: This calculation relies on the given information that PSQR forms a square, implying $\angle SPQ = 90^\circ$ and $PS=PQ$. We have not used any specific properties of the ellipse other than the existence of such a point P, focus S, and directrix point Q on the tangent. The problem statement implies that such a configuration is possible on an ellipse.