Q94 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

For the parabola $y=\frac{h^3}{3} x^2+\frac{h^2}{2} x-h+\frac{3}{4 h^3}$ if the equation of directrix is $y=k$, then $k: h$

  • A
    $16: 19$
  • Correct answer
    B
    $-19: 16$
  • C
    $20: 27$
  • D
    $-27: 20$

Solution

Given, $\quad y=\frac{h^3}{3} x^2+\frac{h^2}{2} x-h+\frac{3}{4 h^3}$ $12 h^3 y=4 h^6 x^2+6 h^5 x-12 h^4+9$ $4 h^6 x^2+6 h^5 x=12 h^3 y+12 h^4-9$ $\left(x^2+\frac{3}{2 h} x\right)=\frac{12 h^4}{4 h^6}\left(y+\frac{12 h^4}{12 h^3}-\frac{3}{14 h^3}\right)$ $\left(x^2+\frac{3}{2 h^2}+\frac{3}{16 h^2}\right)=\frac{3}{h^3}\left(y+h-\frac{3}{4 h^3}\right)+\frac{9}{16 h^2}$ $\left(x+\frac{3}{2 h}\right)^2=\frac{3}{h^3}\left(y+\frac{19 h}{16}-\frac{3}{4 h^3}\right)$ Here, $a=\frac{3}{4 h^3}$ $\therefore$ Equation of directrix is $y+\frac{19 h}{16}-\frac{3}{4 h^3}=\frac{-3}{4 h^3} \Rightarrow y=-\frac{19 h}{16}$ $\therefore \quad k=-\frac{19 h}{16} \Rightarrow \frac{k}{h}=-\frac{19}{16}$ $\therefore \quad k: h=-19: 16$

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

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