Find the equation of the tangent to the curve | vedclass

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

Question

(A) On the $x$-axis,the value of $y = 0$. Setting $y = 0$ in the curve equation $\frac{x-7}{(x-2)(x-3)} = 0$,we get $x = 7$. Thus,the point of interse

Vedclass · Accepted Answer

$20y - x + 7 = 0$

Vedclass · Answer

(A) On the $x$-axis,the value of $y = 0$. Setting $y = 0$ in the curve equation $\frac{x-7}{(x-2)(x-3)} = 0$,we get $x = 7$. Thus,the point of intersection is $(7, 0)$.To find the slope of the tangent,we differentiate $y = \frac{x-7}{x^2 - 5x + 6}$ with respect to $x$ using the quotient rule:$\frac{dy}{dx} = \frac{(x^2 - 5x + 6)(1) - (x-7)(2x-5)}{(x^2 - 5x + 6)^2}$.At the point $(7, 0)$,the denominator is $(7^2 - 5(7) + 6)^2 = (49 - 35 + 6)^2 = (20)^2 = 400$.The numerator at $x=7$ is $(49 - 35 + 6) - (7-7)(2(7)-5) = 20 - 0 = 20$.Thus,the slope $m = \frac{dy}{dx} \Big|_{(7,0)} = \frac{20}{400} = \frac{1}{20}$.The equation of the tangent line at $(7, 0)$ with slope $m = \frac{1}{20}$ is given by $y - y_1 = m(x - x_1)$:$y - 0 = \frac{1}{20}(x - 7)$,$20y = x - 7$,$20y - x + 7 = 0$.

Find the equation of the tangent to the curve $y = \frac{x-7}{(x-2)(x-3)}$ at the point where it cuts the $x$-axis.

Similar Questions

For the curve $4x^{5} = 5y^{4}$,the ratio of the cube of the subtangent at a point on the curve to the square of the subnormal at the same point is

The equation of a straight line passing through the point $(3, 6)$ and cutting the curve $y = \sqrt{x}$ orthogonally is

The chord of the curve $y=x^{2}+2ax+b$ joining the points where $x=\alpha$ and $x=\beta$ is parallel to the tangent to the curve at abscissa $x$ equal to:

If the length of the subnormal at any point on the curve $y = a^{1-n}x^n$ is constant,then $n = \dots$

The equation of the normal to the curve $y = \sin \left( \frac{\pi x}{2} \right)$ at $(1, 1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module