Which of the following is/are true?

Moderate

A

There are infinite positive integral values of a for which $(13 x - 1)^{2} + (13 y - 2)^{2} = {(\frac{5 x + 12 y - 1}{a})}^{2}$ represents an ellipse.
B

The minimum distance of a point (1,2) from the ellipse $4 x^{2} + 9 y^{2} + 8 x - 36 y + 4 = 0 is 1$
C

If from a point P(0, $α$ ) two normals other than the axes are drawn to the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ , then $|α|$ < 9/4.
D

If the length of the latus rectum of an ellipse is one-third of its major axis, then its eccentricity is equal to $\frac{1}{\sqrt{3}}$

Solution

(1) The given equation is
${(x - \frac{1}{13})}^{2} + {(y - \frac{2}{13})}^{2} = \frac{1}{a^{2}} {(\frac{5 x + 12 y - 1}{13})}^{2}$
It represents an ellipse if
$\frac{1}{a^{2}} < 1 or a^{2} > 1 or a > 1$
$(2) 4 x^{2} + 8 x + 9 y^{2} - 36 y = - 4 or 4 (x^{2} + 2 x + 1) + 9 (y^{2} - 4 y + 4) = 36$
$or \frac{(x + 1)^{2}}{9} + \frac{(y - 2)^{2}}{4} = 1$
Hence, (-1, 2) is centre and (1, 2) lies on the major axis. Then the required minimum distance is 1.
(3) The equation of normal at $P (θ) is 5 \sec θx - 4 cosec θ y = 25 - 16$ , and it passes through p(0, $α$ ). Therefore,
$\begin{array}{l} α = \frac{- 9}{4 cosec θ} \\ = \frac{- 9}{4} \sin θ \\ or | α | < \frac{9}{4} \end{array}$
(4) $\frac{2 b^{2}}{a} = \frac{2 a}{3}$
or 3b² = a²
$From b^{2} = a^{2} (1 - e^{2}), we have 1 = 3 (1 - e^{2}) or e = \sqrt{2 / 3} .$

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