If 5x - 2y + k = 0 is a tangent to the parabo | vedclass

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

Question

(C) Given the parabola $y^2 = 6x$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 6$,which implies $\frac{dy}{dx} = \frac{3}{y}$. Give

Vedclass · Accepted Answer

$(\frac{6}{25}, \frac{6}{5})$

Vedclass · Answer

(C) Given the parabola $y^2 = 6x$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 6$,which implies $\frac{dy}{dx} = \frac{3}{y}$. Given the tangent line $5x - 2y + k = 0$,we can rewrite it as $y = \frac{5}{2}x + \frac{k}{2}$. The slope of this line is $m = \frac{5}{2}$. At the point of contact,the slope of the tangent to the parabola must equal the slope of the line: $\frac{3}{y} = \frac{5}{2}$. Solving for $y$,we get $y = \frac{6}{5}$. Substituting $y = \frac{6}{5}$ into the parabola equation $y^2 = 6x$: $(\frac{6}{5})^2 = 6x$ $\frac{36}{25} = 6x$ $x = \frac{36}{25 	imes 6} = \frac{6}{25}$. Thus,the point of contact is $(\frac{6}{25}, \frac{6}{5})$.

If $5x - 2y + k = 0$ is a tangent to the parabola $y^2 = 6x$,then their point of contact is

Similar Questions

If $P$ is a point on the parabola $y=x^{2}+4$ which is closest to the straight line $y =4 x -1,$ then the coordinates of $P$ are:

Tangents are drawn at three points $P(t_1), Q(t_2), R(t_3)$ on the parabola $y^2 = x$. Let these tangents intersect each other at the points $L, M, N$. If $t_1 = 2, t_2 = -4, t_3 = 6$,then the area of the triangle $LMN$ is

The point at which the line $y = mx + c$ touches the parabola $y^2 = 4ax$ is

Two parabolas have the same focus $(4, 3)$ and their directrices are the $x$-axis and the $y$-axis,respectively. If these parabolas intersect at the points $A$ and $B$,then $(AB)^2$ is equal to

The locus of the point of intersection of the perpendicular tangents to the curve $y^2 + 4y - 6x - 2 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module