Minimum distance between two points P and Q on the ellipse x^2 25 + y^2 4 = 1 , if difference between eccentric angles of P and Q is 3 2 , is

Minimum distance between two points P and Q on the ellipse x^2 25…

MCQ

Minimum distance between two points $P$ and $Q$ on the ellipse $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{4} = 1$ , if difference between eccentric angles of $P$ and $Q$ is $\frac{{3\pi }}{2}$ , is

✓
$2\sqrt 2 $
B
$2\sqrt 5 $
C
$\sqrt {29} $
D
$\sqrt {62} $

✓

Answer

Correct option: A.

$2\sqrt 2 $

a
$\mathrm{P}(5 \cos \theta, 2 \sin \theta), \mathrm{Q}(5 \sin \theta,-2 \cos \theta)$

$\mathrm{PQ}=\sqrt{(5 \cos \theta-5 \sin \theta)^{2}+(2 \sin \theta+2 \cos \theta)^{2}}$

$\mathrm{PQ}=\sqrt{29-21 \sin 2 \theta}$

Minmum value of $\mathrm{PQ}=\sqrt{8}=2 \sqrt{2}$

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Match the terms given in Column-I with the terms given in Column-II and choose the correct option from the codes given below.

	Column-I		Column-II
(A)	If $\text{P}(\text{n},4)=20.\text{P}(\text{n},2)$ then the value of n is	(1)	28
(B)	$\ ^5\text{p}_\text{r}=\ ^{26}\text{p}_\text{r-1}$	(2)	4
(C)	$\ ^5\text{p}_\text{r}=\ ^{6}\text{p}_\text{r-1}$	(3)	7
(D)	Value of $\frac{8!}{6!\times2!}$ is	(4)	3