Q110 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics

The ratio between the length of subtangent at any point other than origin on the parabola $y^2=16 a x$ and the abscissa of that point is

  • A
    $1: 3$
  • B
    $1: 4$
  • C
    $1: 2$
  • Correct answer
    D
    $2: 1$

Solution

Let $\left(x_1, y_1\right)$ be any point other than origin on the parabola $y^2=16 a x$. Then, length of subtangent at $$ \begin{aligned} \left(x_1, y_1\right)= & y_1\left(\frac{d x}{d y}\right)_{\left(x_1, y_1\right)} \\ \frac{d y}{d x}= & \frac{16 a}{2 y}=\frac{8 a}{y} \\ & {\left[\because y^2=16 a x \Rightarrow 2 y d y / d x=16 a\right] } \end{aligned} $$ Here, $$ \begin{aligned} \frac{d y}{d x}=\frac{16 a}{2 y} & =\frac{8 a}{y} \\ & {\left[\because y^2=16 a x \Rightarrow 2 y d y / d x=16 a\right] } \end{aligned} $$ $\therefore$ Length of subtangent at $$ \left(x_1, y_1\right)=y_1\left(\frac{y_1}{8 a}\right)=\frac{y_1^2}{8 a} $$ Now, the required ratio $$ =\frac{\text { Length of subtangent at }\left(x_1, \mathrm{y}_1\right)}{x_1}=\frac{y_1^2}{\frac{8 a}{x_1}} $$ $$ \begin{aligned} & =\frac{y_1^2}{8 a} \times \frac{1}{x_1}=\frac{y_1^2}{8 a} \times \frac{16 a}{y_1^2} \\ & \\ & =\frac{2}{1} \quad\left[\because\left(x_1, y_1\right) \text { lies on } y^2=16 a x ; \therefore y_1^2=16 a x_1\right] \end{aligned} $$

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

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