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

Which one of the following curves cuts the parabola $y^2 = 4ax$ at right angles?

The equation of any normal to the parabola ${y^2} = 4a(x - a)$ is

What is the equation of the directrix of the parabola $y^2 + 4y + 4x + 2 = 0$?

Let $P_1 : y = -x^2 + 4x + 2$ and $P_2 : x^2 + 5x + \frac{17}{8} = y$ be two parabolas. Then,the number of common tangents of $P_1$ and $P_2$ is:

The tangents drawn at the endpoints of the latus rectum of a parabola $S = 0$ intersect on the line $x + y = 2$. If $(3, 2)$ is the focus of the parabola,then the axis of the parabola $S = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module