A tangent and a normal are drawn at the point | vedclass

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

Question

(A) The equation of the parabola is $y^{2} = 8x$,so $4a = 8 \Rightarrow a = 2$. The directrix is $x = -a = -2$.$1$. Equation of the tangent at $P(2, -

Vedclass · Accepted Answer

$-16$

Vedclass · Answer

(A) The equation of the parabola is $y^{2} = 8x$,so $4a = 8 \Rightarrow a = 2$. The directrix is $x = -a = -2$.$1$. Equation of the tangent at $P(2, -4)$:Using $T = 0$,we have $y(-4) = 4(x + 2)$ $\Rightarrow -4y = 4x + 8$ $\Rightarrow x + y + 2 = 0$.Intersection with directrix $x = -2$: $-2 + y + 2 = 0 \Rightarrow y = 0$. So,$A(-2, 0)$.$2$. Equation of the normal at $P(2, -4)$:The slope of the tangent is $m_{T} = -1$. The slope of the normal is $m_{N} = -1 / m_{T} = 1$.Equation: $y - (-4) = 1(x - 2)$ $\Rightarrow y + 4 = x - 2$ $\Rightarrow x - y - 6 = 0$.Intersection with directrix $x = -2$: $-2 - y - 6 = 0 \Rightarrow y = -8$. So,$B(-2, -8)$.$3$. Since $AQBP$ is a square,the diagonals $AB$ and $PQ$ bisect each other at the same midpoint $M$.Midpoint of $AB = ((-2 + -2) / 2, (0 + -8) / 2) = (-2, -4)$.Midpoint of $PQ = ((a + 2) / 2, (b - 4) / 2) = (-2, -4)$.Equating coordinates:$(a + 2) / 2 = -2$ $\Rightarrow a + 2 = -4$ $\Rightarrow a = -6$.$(b - 4) / 2 = -4$ $\Rightarrow b - 4 = -8$ $\Rightarrow b = -4$.$4$. Calculate $2a + b$:$2(-6) + (-4) = -12 - 4 = -16$.

$A$ tangent and a normal are drawn at the point $P(2, -4)$ on the parabola $y^{2} = 8x$,which meet the directrix of the parabola at the points $A$ and $B$ respectively. If $Q(a, b)$ is a point such that $AQBP$ is a square,then $2a + b$ is equal to:

Similar Questions

If one end of a focal chord of the parabola $y^2 = 16x$ is at $(1, 4)$,then the length of this focal chord is

$A$ point $P$ moves such that the sum of the angles which the three normals drawn from $P$ to the standard parabola $y^2 = 4ax$ make with the axis of the parabola is constant. Then the locus of $P$ is:

The slope of the chord of the parabola $y^2 = 4ax$ which passes through the origin and is normal at one of its endpoints is:

Find the coordinates of the focus,axis of the parabola,the equation of the directrix,and the length of the latus rectum for $x^{2} = -16y$.

The equation of the latus rectum of the parabola represented by the equation ${y^2} + 2Ax + 2By + C = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module