TS EAMCET 9 May 2024 Shift 2 Solved Paper

Given, ellipse 3x2+4y2=12
⇒‌

x2

4

+‌

y2

3

=1
e=√1−‌

3

4

=√‌

1

4

=‌

1

2

Latus rectums x=±ae=±(2)(‌

1

2

)=±1
∵L1′ is the end of a latus rectum and it is lying in the third quadrant
∵‌‌L1′=(−1,−√3(1−‌

(−1)2

4

))=(−1,−‌

3

2

)
Equation tangent at (−1,‌

−3

2

) is given by
‌

x(−1)

4

+‌

y(− 3 2 )

3

=1⇒x+2y+4=0
Slope of tangent =−‌

1

2

∴ Slope of normal =−(‌

1

−‌ 1 2

)=2. Equation of normal from (−1,−‌

3

2

) is given by
y+‌

3

2

=2(x+1)⇒4x−2y+1=0
The points of intersection of the normal
4x−2y+1=0
and the ellipse ‌

x2

4

+‌

y2

3

=1 are
L1′(−1,−‌

3

2

)‌ and ‌P(‌

11

19

,‌

63

38

)
Hence, a=‌

11

19