TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Given equation of parabola
x2=108y(4a=108⇒a=27)
Equation of tangent, y=mx−27m2 . . . (i)
And equation of parabola
y2=32x(4a=32⇒a=8)
Equation of tangent, y=mx+‌

8

m

. . . (ii)
Since Eqs. (i) and (ii) are identical
∴mx−27m2=mx+‌

8

m

⇒−27m2=‌

8

m

⇒m3=‌

−8

27

⇒m=‌

−2

3

∴ Required equation of tangent
y=(‌

−2

3

)x−27(‌

−2

3

)2
⇒y‌‌=‌

−2x

3

−12
⇒3y‌‌=−2x−36
⇒2x+3y+36‌‌=0