An equation for the line that passes through | vedclass

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

Question

(D) Given the parabola $y = \frac{x^2}{4} - 2$,we have $4y = x^2 - 8$.Differentiating with respect to $x$,we get $4 \frac{dy}{dx} = 2x$,so $\frac{dy}{

Vedclass · Accepted Answer

$x + 2y = 8$

Vedclass · Answer

(D) Given the parabola $y = \frac{x^2}{4} - 2$,we have $4y = x^2 - 8$.Differentiating with respect to $x$,we get $4 \frac{dy}{dx} = 2x$,so $\frac{dy}{dx} = \frac{x}{2}$.Let the point of contact be $P(x_1, y_1)$. The slope of the tangent at $P$ is $m_t = \frac{x_1}{2}$.The slope of the normal at $P$ is $m_n = -\frac{1}{m_t} = -\frac{2}{x_1}$.The line passes through $A(10, -1)$ and $P(x_1, y_1)$,so its slope is $\frac{y_1 - (-1)}{x_1 - 10} = \frac{y_1 + 1}{x_1 - 10}$.Since this line is the normal,$\frac{y_1 + 1}{x_1 - 10} = -\frac{2}{x_1}$.$x_1(y_1 + 1) = -2(x_1 - 10)$ $\Rightarrow x_1y_1 + x_1 = -2x_1 + 20$ $\Rightarrow x_1y_1 + 3x_1 = 20$.Substituting $y_1 = \frac{x_1^2}{4} - 2$,we get $x_1(\frac{x_1^2}{4} - 2) + 3x_1 = 20$.$\frac{x_1^3}{4} - 2x_1 + 3x_1 = 20$ $\Rightarrow \frac{x_1^3}{4} + x_1 = 20$ $\Rightarrow x_1^3 + 4x_1 - 80 = 0$.By inspection,$x_1 = 4$ is a root: $64 + 16 - 80 = 0$.For $x_1 = 4$,$y_1 = \frac{4^2}{4} - 2 = 4 - 2 = 2$. So $P = (4, 2)$.The slope of the normal is $m_n = -\frac{2}{4} = -\frac{1}{2}$.The equation of the line passing through $(10, -1)$ with slope $-\frac{1}{2}$ is $y - (-1) = -\frac{1}{2}(x - 10)$.$2(y + 1) = -(x - 10)$ $\Rightarrow 2y + 2 = -x + 10$ $\Rightarrow x + 2y = 8$.

An equation for the line that passes through $(10, -1)$ and is perpendicular to $y = \frac{x^2}{4} - 2$ is

Similar Questions

If the normal to the ellipse $\frac{x^2}{16} + \frac{y^2}{3} = 1$ at the point $\left( 2, \frac{3}{2} \right)$ is tangent to a parabola,find the equation of the parabola.

Two mutually perpendicular tangents of the parabola $y^2 = 4ax$ meet the axis in $P_1$ and $P_2$. If $S$ is the focus of the parabola,then $\frac{1}{SP_1} + \frac{1}{SP_2}$ is equal to

Find the area of the region enclosed by the equation $2|x| + 3|y| = 6$ in square units.

How many parabolas can be drawn if the endpoints of the latus rectum are given?

Let $\lim_{x \to 2} \frac{(\tan(x-2))(rx^2 + (p-2)x - 2p)}{(x-2)^2} = 5$ for some $r, p \in R$. If the set of all possible values of $q$, such that the roots of the equation $rx^2 - px + q = 0$ lie in $(0, 2)$, be the interval $(\alpha, \beta]$, then $4(\alpha + \beta)$ equals :

Vedclass Test Series

Exam Paper Generator

Online Exam Module