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 to the curve $y = f(x) = x \log_{e} x$ $(x > 0)$ at a point $(c, f(c))$ is parallel to the line segment joining the points $(1, 0)$ and $(e, e)$,then $c$ is equal to:

The equation of the tangent to the curve $y=4 x e^{x}$ at $\left(-1, -\frac{4}{e}\right)$ is

Find the equations of the tangent and the normal to the curve $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$ at the point $(1,3)$.

The $x-$intercept of the tangent at any arbitrary point of the curve $\frac{a}{x^2} + \frac{b}{y^2} = 1$ is proportional to

The angle of intersection of the curves $r = \sin \theta + \cos \theta$ and $r = 2 \sin \theta$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module