What is the equation of the normal to the cur | vedclass

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

Question

(B) Given the curve $y^2 = 4ax$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $\frac{dy}{dx} = \frac{2a}{y}$. At t

Vedclass · Accepted Answer

$x + y - 3a = 0$

Vedclass · Answer

(B) Given the curve $y^2 = 4ax$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $\frac{dy}{dx} = \frac{2a}{y}$. At the point $(a, 2a)$,the slope of the tangent $m_t = \frac{2a}{2a} = 1$. The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\frac{1}{1} = -1$. The equation of the normal at $(a, 2a)$ is $y - y_1 = m_n(x - x_1)$. Substituting the values,we get $y - 2a = -1(x - a)$. $y - 2a = -x + a$. Therefore,$x + y - 3a = 0$.

What is the equation of the normal to the curve $y^2 = 4ax$ at the point $(a, 2a)$?

Similar Questions

$PQ$ is a focal chord of the parabola $y^2 = 4x$ with focus $S$. If $P = (4, 4)$,then $SQ = $

If $b$ and $c$ are the lengths of the segments of any focal chord of the parabola $y^2 = 4ax$,then what is the length of the semi-latus rectum?

The curve described parametrically by $x = t^2 - 2t + 2$ and $y = t^2 + 2t + 2$ represents:

If $y=mx+1$ is a tangent to the parabola $y^2=4x$,then $m=$

If the tangent to the parabola $4y^2 = x$ makes an angle of $60^{\circ}$ with the $x$-axis,then find its point of contact.

Vedclass Test Series

Exam Paper Generator

Online Exam Module