TG EAMCET 2 May 2025 Shift 2 Paper

Given, equations of parabola
‌y2=12x
∴‌‌4a=12⇒a=3
∴ Parametric equation of parabola
x=3t3,y=2(3)t=6t
Any point on the parabola is P(3t2,6t). and focus of parabola is F(3,0) Let Q(x,y) be the point that divides the line segment FP in the ratio m:n
‌∴‌‌x=‌

m(3t2)+n(3)

m+n

=‌

3mt2+3n

m+n

‌‌ And ‌‌‌y=‌

m(6t)+n(0)

m+n

=‌

6mt

m+n

‌⇒‌‌y(m+n)=6mt
‌⇒‌‌t=‌

y(m+n)

6m

‌∴‌‌x=‌

3m( y(m+n) 6m )2+3n

m+n

‌⇒‌‌x=‌

‌ y2(m+n)2 12m +3n

m+n

‌⇒‌‌x=‌

36mn+y2(m+n)2

12m(m+n)

‌⇒‌‌12m(m+n)x=36mn+y2(m+n)2
‌⇒‌‌y2(m+n)2=12m(m+n)x−36mn
‌⇒‌‌y2=‌

12m(m+n)

(m+n)

x−‌

36mn

(m+n)2

‌⇒‌‌y2=‌

12m

m+n

x−‌

36mn

(m+n)2

Which is in the form of Y2=4AX+B
∴‌ Length of latusrectum ‌=‌

12m

m+n