Manipal Entrance Test 2020 Math Paper

Given equation is
b2x2+y2=a2b2
⇒‌‌‌

x2

a2

+‌

y2

a2b2

=1......(i)
Above equation of an ellipse with semi-major axis (a) and semi-minor axis (ab).
Now, ecentricity,
e=1−‌

a2b2

a2

⇒b2=1−e2......(ii)
Let (x,y) be extrimities of latusrectum, then
x=ae and y=±‌

a2b2

a

⇒‌‌‌

x

a

=e and ‌

y

a

=±b2
From Eq. (ii), we qet
±‌

y

a

=1−‌

x2

a2

⇒±ay+x2=a2
Hence, locus of latusrectum is x2±ay=a2.