The equation of the normal to the parabola x^ | vedclass

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

Question

(D) Given the parabola equation is $x^2 = 8y$. When $x = 4$,we substitute into the equation: $4^2 = 8y \Rightarrow 16 = 8y \Rightarrow y = 2$. So,the

Vedclass · Accepted Answer

$x + y = 6$

Vedclass · Answer

(D) Given the parabola equation is $x^2 = 8y$. When $x = 4$,we substitute into the equation: $4^2 = 8y \Rightarrow 16 = 8y \Rightarrow y = 2$. So,the point of contact is $(4, 2)$. Differentiating $x^2 = 8y$ with respect to $x$,we get $2x = 8 \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{2x}{8} = \frac{x}{4}$. The slope of the tangent at $x = 4$ is $m_t = \left. \frac{x}{4} \right|_{x=4} = 1$. The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\frac{1}{1} = -1$. The equation of the normal at $(4, 2)$ is $y - y_1 = m_n(x - x_1)$. Substituting the values: $y - 2 = -1(x - 4)$. $y - 2 = -x + 4$. $x + y = 6$.

The equation of the normal to the parabola $x^2 = 8y$ at $x = 4$ is

Similar Questions

What is the equation of the directrix of the parabola $x^2 = -8y$?

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

The length of the latus rectum of the parabola $2\{(x - 1)^2 + (y - 3)^2\} = (x + y - 1)^2$ is:

If the locus of a point which divides a chord with slope $2$ of the parabola $y^2=4x$ internally in the ratio $1:3$ is a parabola,then its vertex is

If the axis of a parabola is horizontal and it passes through the points $(0, 0), (0, -1)$ and $(6, 1)$,then its equation is

Vedclass Test Series

Exam Paper Generator

Online Exam Module