TG EAMCET 2 May 2025 Shift 1 Paper

Let the equation of the line L be ‌

x

a

+‌

y

b

=1, where a and b are the x and y intercepts, respectively. So, OA=a and OB=b,a>0,b>0.
Here, the line passes through the point (4,9), So,
‌

4

a

+‌

9

b

=1
We want to minimize a+b. So consider
(a+b)(‌

4

a

+‌

9

b

)‌=4+‌

9a

b

+‌

4b

a

+9
‌=13+‌

9a

b

+‌

4b

a

By AM-GM inequality, for positive numbers,
‌‌

9a

b

+‌

4b

a

≥‌2√‌

9a

b

⋅‌

4b

a

⇒2√36=2⋅6=12
So, 13+‌

9a

b

+‌

4b

a

=13+12=25
But, since (a+b)(‌

4

a

+‌

9

b

)=(a+b)⋅1
=a+b‌‌‌ (Using Eq. (i)) ‌
∴ The minimum value of a+b is 25 and this occurs when ‌

9a

b

=‌

4b

a

⇒‌9a2‌=4b2‌ or ‌3a=2b‌‌(∵a,b>0)
⇒‌b‌=‌

3

2

a
So, ‌

4

a

+‌

9

b

=1
‌⇒‌

4

a

+‌

9

3 2 a

=1⇒‌

4

a

+‌

18

3a

=1
‌⇒‌

4

a

+‌

6

a

=1⇒‌

10

a

=1
‌⇒a=10
Then, b=‌

3

2

a=‌

3

2

×10=15
So, the minimum value of OA+OB is a+b=10+15=25