Let P1 be a parabola with vertex (3, 2) and focus (4, 4) and P2 be its mirror image with respect to the line x + 2y = 6. Then the directrix of P2 is x + 2y = _______.
The correct answer is 10
Focus = (4, 4) and vertex = (3, 2)
∴ Point of intersection of directrix with axis of parabola = A = (2, 0)
Image of A(2, 0) with respect to line
x + 2y = 6 is B(x2, y2)
\(∴\frac{x^2−2}{1}=\frac{y^2−0}{2}=\frac{−2(2+0−6)}{5}\)
\(∴B(x_2,y_2)=(185,165).\)
Point B is point of intersection of direction with axes of parabola P2.
∴ x + 2y = λ must have point
\((\frac{18}{5},\frac{16}{5})\)
∴ x + 2y = 10
If \( x^2 = -16y \) is an equation of a parabola, then:
(A) Directrix is \( y = 4 \)
(B) Directrix is \( x = 4 \)
(C) Co-ordinates of focus are \( (0, -4) \)
(D) Co-ordinates of focus are \( (-4, 0) \)
(E) Length of latus rectum is 16
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:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the river bank is _________ cm. (Take \( g = 10 \, {m/s}^2 \)).
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
=> MP2 = PS2
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2