Match the following lists and then choosen the corrct code.

Options

A. 1) `{:(a,b,c,d),(q,p,s,r):}`
B. 2) `{:(a,b,c,d),(q,r,s,q):}`
C. 3) `{:(a,b,c,d),(s,p,q,r):}`
D. 4) `{:(a,b,c,d),(q,q,s,p):}`

Explanation

(a) Let one of the vertices of the rectangle be
`P(a cos theta, b sin theta)`. Then its area is
`A=(2 a cos theta)(2b sin theta)=2ab sin 2theta`
Hence, `A_("max")=2ab`
Now, the area of recrtangle formed by the extermities of the latus rectum is
`2aexx(2b^(2))/(a)=4eb^(2)`
Given that
`2ab=4eb^(2)`
or `(2b)/(a)e=1`
`or (4b^(2))/(a^(2))e^(2)=1`
or `4(1-e^(2))e^(2)=1`
`or(2e^(2)-1)^(2)=0`
(b)
For the ellipse, the distacec between the foci, `2ae=8`, and the length of the semi-minor axis is b=4 . Now,
`b^(2)=a^(2)-a^(2)e^(2)`
or `16=a^(2)-16`
`or e^(2)=32`
or `e=sqrt(1-(16)/(32))=(1)/(sqrt(2))`
(c) Normal at point P(6,2) to the ellipse passes through its focus Q(5,2). Then P must be the extermity of the major axis, Now, ae=QR=1 (where R is ceter ) and a-ae=1

`:. a=2`
`b^(2)=a^(2)-a^(2)e^(2)=4-1=3`
or `e=sqrt(1-(3)/(4))=(1)/(2)`
(d)
The extermities of the lotus rectum of teh parabola `y^(2)=24x` are `(6,+-12)`. For ellipse `2be=24` and the extermity of the minro axis is (0,0). Hence a=6. Now,
`a^(2)=b^(2)-b^(2)e^(2)`
or `b^(2)=36+144=180`
or `e^(2)sqrt(1-(36)/(180))=sqrt(1-(1)/(5)) =(2)/(sqrt(5))`