TS EAMCET 04 May 2019 Shift 1 Solved Paper

It is given that x+y+n=0 is normal to the ellipse x2+3y2=3
this implies,
n‌‌=±‌

(3−1)

√3+1

‌‌=±1
since n cannot be negative then,
n=1
Consider the equation of normal,
x+y+1=0‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I)
The equation x+my+3=0 is tangent of ellipse x2+5y2=5
so,
‌

3

m

‌‌=±√‌

5

m2

+1
‌

9

m2

‌‌=‌

5

m2

+1
m2‌‌=4
m‌‌=±2
The m cannot be greater than zerothen,
m=−2
Consider equation of tangent,
x−2y+3=0‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(II)
Solve equation (I) and (II),
x=−‌

5

3

,y=‌

2

3

The above values satisfy the equation,
x−5y+5=0
Therefore, the correct option is (2)