If the tangent and the normal drawn to the cu | vedclass

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

Question

(D) Given the curve $xy^2 + x^2y = 12$. Differentiating with respect to $x$: $y^2 + 2xy \frac{dy}{dx} + 2xy + x^2 \frac{dy}{dx} = 0$. At the point $(1

Vedclass · Accepted Answer

$\frac{274}{35}$

Vedclass · Answer

(D) Given the curve $xy^2 + x^2y = 12$. Differentiating with respect to $x$: $y^2 + 2xy \frac{dy}{dx} + 2xy + x^2 \frac{dy}{dx} = 0$. At the point $(1, 3)$,we have $3^2 + 2(1)(3) \frac{dy}{dx} + 2(1)(3) + (1)^2 \frac{dy}{dx} = 0$. $9 + 6 \frac{dy}{dx} + 6 + \frac{dy}{dx} = 0 \implies 7 \frac{dy}{dx} = -15 \implies \frac{dy}{dx} = -\frac{15}{7}$. The equation of the tangent at $(1, 3)$ is $y - 3 = -\frac{15}{7}(x - 1)$. For $T$,set $y = 0$: $-3 = -\frac{15}{7}(x - 1) \implies 21 = 15(x - 1) \implies x - 1 = \frac{21}{15} = \frac{7}{5} \implies x = 1 + \frac{7}{5} = \frac{12}{5}$. So $T = (\frac{12}{5}, 0)$. The equation of the normal at $(1, 3)$ is $y - 3 = \frac{7}{15}(x - 1)$. For $N$,set $y = 0$: $-3 = \frac{7}{15}(x - 1) \implies -45 = 7(x - 1) \implies x - 1 = -\frac{45}{7} \implies x = 1 - \frac{45}{7} = -\frac{38}{7}$. So $N = (-\frac{38}{7}, 0)$. $TN = |\frac{12}{5} - (-\frac{38}{7})| = |\frac{12}{5} + \frac{38}{7}| = |\frac{84 + 190}{35}| = \frac{274}{35}$.

If the tangent and the normal drawn to the curve $xy^2 + x^2y = 12$ at the point $(1, 3)$ meet the $X$-axis in $T$ and $N$ respectively,then $TN =$

Similar Questions

If the tangent at any point on the curve $x^4 + y^4 = a^4$ meets the axes at $p$ and $q$,then the value of $p^{-4/3} + q^{-4/3}$ is:

If the normal to the curve $y = f(x)$ at the point $(3, 4)$ makes an angle of $3\pi / 4$ with the positive $X$-axis,find $f'(3)$.

If the tangent to the curve $xy+ax+by=0$ at $(1,1)$ makes an angle $\tan^{-1} 2$ with the $x$-axis,then $\frac{a+b}{ab} =$

If the normal drawn at the point $P$ on the curve $y=x \log x$ is parallel to the line $2x-2y=3$,then $P=$

If the normal to the curve $y = f(x)$ at the point $(4, 6)$ makes an angle $\frac{2\pi}{3}$ with the positive $x$-axis in the anticlockwise direction,then $f'(4)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module