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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(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

Similar Questions

The line $x + y = 0$ touches the curve $y^2 = ax^3 + b$ at the point $(1, -1)$. Find the values of $a$ and $b$.

The equation of the normal to the curve $y(1+x^{2})=2-x$ at the point where the tangent crosses the $x$-axis is

What is the equation of the tangent to the curve $y = 2 \cos x$ at $x = \pi / 4$?

Let $\Gamma$ be the curve $y=b e^{-x/a}$ and $L$ be the straight line $\frac{x}{a}+\frac{y}{b}=1$, where $a, b \in \mathbb{R}$. Which of the following statements is true?

The values of parameter $a$ such that the line $(\log_{2}(1 + 5a - a^{2}))x - 5y - (a^{2} - 5) = 0$ is a normal to the curve $xy = 1$ may lie in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module