Which of the following is/are true ?

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

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The minimum distance of a point

(1, 2)

from the ellipse

4 x^{2} + 9 y^{2} + 8 x - 36 y + 4 = 0

1

unit

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If from a point

P (0, α)

two normals other than axes are drawn to the ellipse

\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1

, then

| α | < \frac{9}{4}

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If the length of latus rectum of an ellipse is one-third of its major axis, then its eccentricity is equal to

\frac{1}{\sqrt{3}}

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Solution

The correct options are
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 $1$ unit
C If from a point $P (0, α)$ two normals other than axes are drawn to the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ , then $| α | < \frac{9}{4}$
$a.$ 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 ellipse if $\frac{1}{a^{2}} < 1 \Rightarrow a^{2} > 1 \Rightarrow a \in (- \infty, - 1) \cup (1, \infty)$
Hence, there are infinite positive integrals possible.

$b. 4 x^{2} + 8 x + 9 y^{2} - 36 y = - 4$
$\Rightarrow 4 (x^{2} + 2 x + 1) + 9 (y^{2} - 4 y + 4) = 36$

$\Rightarrow \frac{(x + 1)^{2}}{9} + \frac{(y - 2)^{2}}{4} = 1$
Hence, $(- 1, 2)$ is the centre and $(1, 2)$ lies on the major axis. Then required minimum distance is $1$ .

$c.$ Equation of normal at $P (θ)$ is $5 sec θ x - 4 cosec θ y = 25 - 16,$ and it passes through $P (0, α)$
$∴ α = - \frac{9}{4 cosec θ}$
$\Rightarrow α = - \frac{9}{4} sin θ$
$\Rightarrow α \in (- \frac{9}{4}, \frac{9}{4})$
because for $sin 90 °$ or $sin 270 °$ $P (θ)$ will become end point of minor axis
$\Rightarrow | α | < \frac{9}{4}$

$d.$
Length of latus-rectum $= \frac{1}{3}$ (length of major axis)
$\frac{2 b^{2}}{a} = \frac{2 a}{3} \Rightarrow 3 b^{2} = a^{2}$
$\Rightarrow$ From $b^{2} = a^{2} (1 - e^{2}),$
$1 = 3 (1 - e^{2})$
$\Rightarrow e = \sqrt{\frac{2}{3}}$

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