Q107 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics
If $\frac{k}{\alpha^3}$ is the length of the sub normal at any point $P(\alpha, y)$ on the curve $x^2-a^2=\frac{x^2 y^2}{a^2}$, then $k=$
- Aa
- B$a^2$
- C$\frac{3 a}{2}$
- Correct answerD$a^4$
Solution
Given curve, $x^2-a^2=\frac{x^2 y^2}{a^2}$
$\Rightarrow \quad 2 x=\frac{1}{a^2}\left[x^2(2 y) \frac{d y}{d x}+(2 x) y^2\right]$
$\Rightarrow \quad 1=\frac{1}{a^2}\left[x y \frac{d y}{d x}+y^2\right]$
$\Rightarrow x y \frac{d y}{d x}+y^2=a^2 \Rightarrow x y \frac{d y}{d x}=a^2-y^2$
$\Rightarrow \quad \frac{d y}{d x}=\frac{a^2-y^2}{x y} \ldots$
$\Rightarrow\left(\frac{d y}{d x}\right)_{P(\alpha, y)}=\frac{a^2-y^2}{\alpha y}$ $\ldots(\mathrm{i})$
Now, Since $P(\alpha, y)$ lies on the curve.
$\therefore \quad \alpha^2-a^2=\frac{\alpha^2 y^2}{a^2} \Rightarrow y^2=\frac{a^2}{\alpha^2}\left(\alpha^2-a^2\right)$ $\ldots(\mathrm{ii})$
$\therefore$ Length of subnormal at $P(\alpha, y)$ is
$y_1\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=y\left(\frac{a^2-y^2}{\alpha y}\right)$(using Eq. (i))
$=\frac{a^2-\left(\frac{a^2}{\alpha 2}\left(\alpha^2-a^2\right)\right)}{\alpha}$ (using Eq.(ii))
$=\frac{a^4}{\alpha^3}=\frac{k}{\alpha^3}$ (given)
$\therefore \quad k=a^4$
Hence, the correct answer is option: 'D'.