Let P_{1} be a parabola with vertex (3,2) and | vedclass

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

Question

(A) The axis of the parabola $P_{1}$ passes through the vertex $(3,2)$ and focus $(4,4)$. The slope of the axis is $m = \frac{4-2}{4-3} = 2$.Since the

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) The axis of the parabola $P_{1}$ passes through the vertex $(3,2)$ and focus $(4,4)$. The slope of the axis is $m = \frac{4-2}{4-3} = 2$.Since the axis is perpendicular to the directrix,the slope of the directrix is $-\frac{1}{2}$.Thus,the equation of the directrix is of the form $x + 2y = k$.The distance from the vertex $(3,2)$ to the directrix is equal to the distance from the vertex to the focus,which is $a = \sqrt{(4-3)^2 + (4-2)^2} = \sqrt{1^2 + 2^2} = \sqrt{5}$.Using the distance formula from point $(3,2)$ to line $x + 2y - k = 0$:$\frac{|3 + 2(2) - k|}{\sqrt{1^2 + 2^2}} = \sqrt{5} \implies |7 - k| = 5$.This gives $7 - k = 5 \implies k = 2$ or $7 - k = -5 \implies k = 12$.Since the focus $(4,4)$ satisfies $4 + 2(4) = 12$,the line $x + 2y = 12$ passes through the focus and cannot be the directrix. Thus,the directrix of $P_{1}$ is $x + 2y = 2$.Let the line of reflection be $L: x + 2y = 6$. The mirror image of the line $x + 2y = 2$ with respect to $x + 2y = 6$ is found by noting that the lines are parallel.If the line $x + 2y = c$ is the image of $x + 2y = 2$ across $x + 2y = 6$,then $6$ is the arithmetic mean of $2$ and $c$:$\frac{2 + c}{2} = 6 \implies 2 + c = 12 \implies c = 10$.Therefore,the directrix of $P_{2}$ is $x + 2y = 10$.

Let $P_{1}$ be a parabola with vertex $(3,2)$ and focus $(4,4)$,and let $P_{2}$ be its mirror image with respect to the line $x + 2y = 6$. Then the directrix of $P_{2}$ is $x + 2y =$

Similar Questions

$P$ and $Q$ are two points on the parabola $y^2 = 8x$ and $S$ is its focus. $PS$ and $QS$ meet the curve again in $T$ and $R$ respectively. If $PQ$ passes through a fixed point $(-2, 3)$,then $TR$ also passes through a fixed point whose coordinates are

If a normal to the parabola $y^2=12x$ at $A(3,-6)$ cuts the parabola again at $P$,then the equation of the tangent at $P$ is

The Cartesian equation of the parabola $x = -2 + 2t^2$,$y = 2 + 4t$ is

The equation of the directrix of the parabola $x^2+8x+12y+4=0$ is

$A$ line passes through the point $(-1, 1)$ and makes an angle $\sin^{-1}(\frac{3}{5})$ in the positive direction of the $x$-axis. If this line meets the curve $x^2 = 4y - 9$ at points $A$ and $B$, then the length $|AB|$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module