Q108 / 160 · AP EAMCET 2019 (21 Apr Shift 1) · Mathematics

The number of tangent to the curve $y^2(x-a)=x^2(x+a)(a>0)$ that are parallel to the $X$-axis is

  • A
    infinitely many
  • Correct answer
    B
    0
  • C
    1
  • D
    2

Solution

Given, curve is $$ y^2(x-a)=x^2(x+a) $$ Apply log on both sides $$ \begin{array}{ll} & \log \left\{y^2(x-a\}=\log \left\{x^2(x+a)\right\}\right. \\ \Rightarrow \quad & 2 \log y+\log (x-a)=2 \log x+\log (x+a) \end{array} $$ Differentiating both sides w.r.t. $x$, we get $$ \Rightarrow \quad \frac{2}{y} \cdot \frac{d y}{d x}+\frac{1}{x-a}=\frac{2}{x}+\frac{1}{x+a} $$ $$ \begin{aligned} & \Rightarrow \quad \frac{2}{y} \frac{d y}{d x}=\frac{2}{x}+\frac{1}{x+a}-\frac{1}{x-a} \\ & =\frac{2(x+a)(x-a)+x(x-a)-x(x+a)}{x(x+a)(x-a)} \\ & =\frac{2\left(x^2-a^2\right)+x^2-a x-x^2-a x}{x(x-a)(x+a)} \\ & \Rightarrow \quad \frac{2}{y} \frac{d y}{d x}=\frac{2 x^2-2 a^2-2 a x}{x(x-a)(x+a)} \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{x^2-a^2-a x}{x(x-a)(x+a)} \cdot x \cdot \sqrt{\frac{x+a}{x-a}} \\ & \text { At } \quad \frac{d y}{d x}=0 \\ & \therefore \quad x^2-a^2-a x=0 \\ & \text { Here, } p=a^2+4 a^2>0 \\ & \end{aligned} $$ So, it has two real roots.

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

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