The equation of the line passing through the | vedclass

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

Question

(A) The equation of a line passing through $(1/2, 2)$ with slope $m$ is $y - 2 = m(x - 1/2)$,which simplifies to $y = mx - m/2 + 2$.For this line to b

Vedclass · Accepted Answer

$2x + 2y - 5 = 0$

Vedclass · Answer

(A) The equation of a line passing through $(1/2, 2)$ with slope $m$ is $y - 2 = m(x - 1/2)$,which simplifies to $y = mx - m/2 + 2$.For this line to be tangent to the parabola $y = -x^2/2 + 2$,we substitute $y$ into the parabola equation:$mx - m/2 + 2 = -x^2/2 + 2$$x^2/2 + mx - m/2 = 0$$x^2 + 2mx - m = 0$For tangency,the discriminant $D = 0$:$D = (2m)^2 - 4(1)(-m) = 0$$4m^2 + 4m = 0$$4m(m + 1) = 0$So,$m = 0$ or $m = -1$.If $m = 0$,the line is $y = 2$,which is a tangent to the parabola at $(0, 2)$ but is not a secant to the curve $y = \sqrt{4 - x^2}$ (it is a tangent to the circle at $(0, 2)$).If $m = -1$,the line is $y - 2 = -1(x - 1/2)$,which simplifies to $y = -x + 5/2$,or $2x + 2y - 5 = 0$. This line intersects the curve $y = \sqrt{4 - x^2}$ at two points,thus it is a secant line.

The equation of the line passing through the point $(1/2, 2)$ and tangent to the parabola $y = -\frac{x^2}{2} + 2$ and secant to the curve $y = \sqrt{4 - x^2}$ is

Similar Questions

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

If $x = my + c$ is a normal to the parabola ${x^2} = 4ay$,then the value of $c$ is

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:

The length of the latus rectum of the conic $25[(x-2)^2+(y-3)^2]=(3x-4y+7)^2$ is

The length of the normal chord to the parabola $y^2 = 4x$,which subtends a right angle at the vertex,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module