The normal to the curve y(x-2)(x-3)=x+6 at th | vedclass

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

Question

(B) The given equation of the curve is $y(x-2)(x-3)=x+6$. At the $Y$-axis,$x=0$. Substituting $x=0$ into the equation: $y(0-2)(0-3)=0+6 \Rightarrow y(

Vedclass · Accepted Answer

$\left(\frac{1}{2}, \frac{1}{2}\right)$

Vedclass · Answer

(B) The given equation of the curve is $y(x-2)(x-3)=x+6$. At the $Y$-axis,$x=0$. Substituting $x=0$ into the equation: $y(0-2)(0-3)=0+6 \Rightarrow y(-2)(-3)=6 \Rightarrow 6y=6 \Rightarrow y=1$. So,the point of intersection is $(0, 1)$. Now,rewrite the equation as $y = \frac{x+6}{x^2-5x+6}$. Differentiating with respect to $x$ using the quotient rule: $\frac{dy}{dx} = \frac{(x^2-5x+6)(1) - (x+6)(2x-5)}{(x^2-5x+6)^2}$. At $x=0$,the slope of the tangent $m_t = \frac{(0-0+6)(1) - (0+6)(0-5)}{(0-0+6)^2} = \frac{6 - (-30)}{36} = \frac{36}{36} = 1$. The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{1} = -1$. The equation of the normal at $(0, 1)$ is $y - 1 = -1(x - 0)$,which simplifies to $y = -x + 1$ or $x + y = 1$. Checking the options,for $(\frac{1}{2}, \frac{1}{2})$,we have $\frac{1}{2} + \frac{1}{2} = 1$,which satisfies the equation. Thus,the normal passes through $(\frac{1}{2}, \frac{1}{2})$.

The normal to the curve $y(x-2)(x-3)=x+6$ at the point,where the curve intersects the $Y$-axis,passes through the point

Similar Questions

If the curves $2x = y^2$ and $2xy = K$ intersect perpendicularly,then the value of $K^2$ is

The angle between the tangents to the curves $y=2x^2$ and $x=2y^2$ at $(1,1)$ is

If $y=2x$ is a tangent to the curve $y^2=ax^3+b$ at $(1,2)$, then $(a, b)=$

The equation of the tangent to the curve $y = \cos(x + y)$ for $-2\pi \leq x \leq 2\pi$,which is parallel to the line $x + 2y = 0$,is:

The length of the subtangent to the curve $\sqrt{x} + \sqrt{y} = 3$ at the point $(4, 1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module