Q95 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

The equation of the common tangent of the parabolas $x^2=108 y$ and $y^2=32 x$ is

  • Correct answer
    A
    $2 x+3 y+36=0$
  • B
    $2 x+3 y=36$
  • C
    $3 x+2 y+36=0$
  • D
    $3 x+2 y=36$

Solution

(a) Given equation of parabola $x^2=108 y(4 a=108 \Rightarrow a=27)$ Equation of tangent, $y=m x-27 m^2$ $\ldots(\mathrm{i})$ And equation of parabola $y^2=32 x(4 a=32 \Rightarrow a=8)$ Equation of tangent, $y=m x+\frac{8}{m}$ $\ldots(\mathrm{ii})$ Since Eqs. (i) and (ii) are identical $m x-27 m^2=m x+\frac{8}{m}$ $\Rightarrow \quad-27 m^2=\frac{8}{m} \Rightarrow m^3=\frac{-8}{27} \Rightarrow m=\frac{-2}{3}$ $\therefore$ Required equation of tangent $y=\left(\frac{-2}{3}\right) x-27\left(\frac{-2}{3}\right)^2$ $\Rightarrow \quad y=\frac{-2 x}{3}-12$ $\Rightarrow \quad 3 y=-2 x-36$ $\Rightarrow \quad 2 x+3 y+36=0$

Hence, the correct answer is option: 'A'.

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