Which of the following is/are true about the ellipse $x^{2} + 4 y^{2} - 2 x - 16 y + 13 = 0 ?$

A

The latus rectum of the ellipse is 1.
B

The distance between the foci of the ellipse is $4 \sqrt{3}$
C

The sum of the focal distances of a point P(x, y) on the ellipse is 4.
D

Line y = 3 meets the tangents drawn at the vertices of the ellipse at points P and Q. Then PQ subtends a right angle at any of its foci.

Solution

$\begin{array}{ll} x^{2} + 4 y^{2} - 2 x - 16 y + 13 = 0 \\ or (x^{2} - 2 x + 1) + 4 (y^{2} - 4 y + 4) = 4 \\ or \frac{(x - 1)^{2}}{4} + \frac{(y - 2)^{2}}{1} = 1 \end{array}$
Length of latus rectum $= \frac{2 \times 1}{2} = 1$
$Also, e = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$
$or 2 ae = 2 \times 2 \times \frac{\sqrt{3}}{2} = 2 \sqrt{3}$
Sum of focal distance = 2a = 4
Tangents at the vertices are
$x - 1 = \pm 2$
or x = 3, -1
Therefore, the line y = 3 intersects these at points P(3, 3) and Q(-1, 3).
One of the foci is $S (\sqrt{3} + 1, 2)$
Slope of $PS = \frac{1}{2 - \sqrt{3}}$
Slope of $QS = \frac{1}{- 2 - \sqrt{3}}$
$∴$ Product of slopes $= \frac{1}{2 - \sqrt{3}} \times \frac{1}{- 2 - \sqrt{3}} = - 1$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME