The number of tangents to the curve y^2(x-a)= | vedclass

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(D) The given curve is $y^2(x-a)=x^2(x+a)$.For a tangent to be parallel to the $X$-axis,the slope $\frac{dy}{dx}$ must be $0$.Differentiating $y^2(x-a

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$2$

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(D) The given curve is $y^2(x-a)=x^2(x+a)$.For a tangent to be parallel to the $X$-axis,the slope $\frac{dy}{dx}$ must be $0$.Differentiating $y^2(x-a)=x^2(x+a)$ with respect to $x$ using implicit differentiation:$2y \frac{dy}{dx}(x-a) + y^2(1) = 2x(x+a) + x^2(1)$Setting $\frac{dy}{dx} = 0$,we get:$y^2 = 2x^2 + 2ax + x^2 = 3x^2 + 2ax$Substitute $y^2 = \frac{x^2(x+a)}{x-a}$ into the equation:$\frac{x^2(x+a)}{x-a} = 3x^2 + 2ax$Since $x=0$ is a solution (giving $y=0$),we check for other solutions by dividing by $x$ $(x 
eq 0)$:$\frac{x(x+a)}{x-a} = 3x + 2a$$x^2 + ax = (3x + 2a)(x - a)$$x^2 + ax = 3x^2 - 3ax + 2ax - 2a^2$$x^2 + ax = 3x^2 - ax - 2a^2$$2x^2 - 2ax - 2a^2 = 0$$x^2 - ax - a^2 = 0$The roots are $x = \frac{a \pm \sqrt{a^2 - 4(1)(-a^2)}}{2} = \frac{a \pm a\sqrt{5}}{2}$.For these values of $x$,$y^2 = 3x^2 + 2ax$. Since $x^2 - ax - a^2 = 0$,$x^2 = ax + a^2$.$y^2 = 3(ax + a^2) + 2ax = 5ax + 3a^2$.For $x = \frac{a(1+\sqrt{5})}{2}$,$y^2 = 5a(\frac{a(1+\sqrt{5})}{2}) + 3a^2 > 0$,giving two values of $y$.For $x = \frac{a(1-\sqrt{5})}{2}$,$y^2 = 5a(\frac{a(1-\sqrt{5})}{2}) + 3a^2 = \frac{5a^2 - 5a^2\sqrt{5} + 6a^2}{2} = \frac{11-5\sqrt{5}}{2}a^2 Thus,there are $2$ tangents.

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

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