The equation of the normal at the point left( | vedclass

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

Question

(B) For the parabola $y^2 = 4ax$,the slope of the tangent at any point $(x_1, y_1)$ is given by $2y y_1 = 4a(x + x_1) \implies y = \frac{2a}{y_1}(x +

Vedclass · Accepted Answer

$4x + 8y - 9a = 0$

Vedclass · Answer

(B) For the parabola $y^2 = 4ax$,the slope of the tangent at any point $(x_1, y_1)$ is given by $2y y_1 = 4a(x + x_1) \implies y = \frac{2a}{y_1}(x + x_1)$.At the point $\left( \frac{a}{4}, a \right)$,the slope of the tangent $m_t = \frac{2a}{a} = 2$.The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{2}$.The equation of the normal at $\left( \frac{a}{4}, a \right)$ is given by $y - y_1 = m_n(x - x_1)$.$y - a = -\frac{1}{2}(x - \frac{a}{4})$.$2(y - a) = -(x - \frac{a}{4})$.$2y - 2a = -x + \frac{a}{4}$.$x + 2y = 2a + \frac{a}{4} = \frac{9a}{4}$.$4x + 8y = 9a$.$4x + 8y - 9a = 0$.

The equation of the normal at the point $\left( \frac{a}{4}, a \right)$ to the parabola $y^2 = 4ax$ is:

Similar Questions

If $x-2y+k=0$ is a tangent to the parabola $y^2-4x-4y+8=0$,then the slope of the tangent drawn at $(1, k)$ on the given parabola is

The equation of the locus of all points equidistant from the point $(4, 2)$ and the $x$-axis is:

The maximum number of normals that can be drawn from a point to a parabola is

If $x-2=t^2$ and $y=2t$ are the parametric equations of the parabola $y^2=a(x-b)$,then the value of $a+b$ equals

In the parabola $y^2 = 6x$,the equation of the chord passing through the vertex and the negative end of the latus rectum is

Vedclass Test Series

Exam Paper Generator

Online Exam Module