The normal to the curve 2x^2 + y^2 = 12 at th | vedclass

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

Question

(B) Given the curve equation $2x^2 + y^2 = 12$. Differentiating with respect to $x$,we get $4x + 2yy' = 0$,which implies $y' = -\frac{2x}{y}$. At the

Vedclass · Accepted Answer

$\left( \frac{-22}{9}, \frac{-2}{9} \right)$

Vedclass · Answer

(B) Given the curve equation $2x^2 + y^2 = 12$. Differentiating with respect to $x$,we get $4x + 2yy' = 0$,which implies $y' = -\frac{2x}{y}$. At the point $(2, 2)$,the slope of the tangent $m_t = -\frac{2(2)}{2} = -2$. The slope of the normal $m_n = -\frac{1}{m_t} = \frac{1}{2}$. The equation of the normal at $(2, 2)$ is $y - 2 = \frac{1}{2}(x - 2)$,which simplifies to $2y - 4 = x - 2$,or $x = 2y - 2$. Substituting $x = 2y - 2$ into the curve equation $2x^2 + y^2 = 12$: $2(2y - 2)^2 + y^2 = 12$ $2(4y^2 - 8y + 4) + y^2 = 12$ $8y^2 - 16y + 8 + y^2 = 12$ $9y^2 - 16y - 4 = 0$ $(9y + 2)(y - 2) = 0$. Thus,$y = 2$ or $y = -\frac{2}{9}$. For $y = 2$,$x = 2(2) - 2 = 2$ (the original point). For $y = -\frac{2}{9}$,$x = 2(-\frac{2}{9}) - 2 = -\frac{4}{9} - \frac{18}{9} = -\frac{22}{9}$. Therefore,the normal meets the curve again at $\left( -\frac{22}{9}, -\frac{2}{9} \right)$.

The normal to the curve $2x^2 + y^2 = 12$ at the point $(2, 2)$ meets the curve again at

Similar Questions

If the tangent at the point $P$ with coordinates $(h, k)$ on the curve $y^{2}=2x^{3}$ is perpendicular to the straight line $4x=3y$, then

The coordinates of the point on the curve $y = x \log x$ at which the normal is parallel to the line $2x - 2y = 3$ are:

For the curve $b{y^2} = {(x + a)^3}$,the square of the subtangent is proportional to:

The length of the tangent drawn at the point $P\left(\frac{\pi}{4}\right)$ on the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=2^{\frac{2}{3}}$ is

If the points of contact of the tangents drawn from $(0,0)$ to the curve $y=x^2+3x+4$ are $(\alpha, \beta)$ and $(\gamma, \delta)$,then $\beta+\delta=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module