To solve the problem of finding the point on the curve y² = 16x where the y-coordinate changes twice as fast as the x-coordinate, we follow the derivative approach:
1. Given the equation of the curve: y² = 16x, differentiate both sides with respect to t (time) to understand how y and x change.
2. Applying implicit differentiation: 2y(dy/dt) = 16(dx/dt).
3. Given that the y-coordinate changes twice as fast as the x-coordinate, we have: dy/dt = 2(dx/dt).
4. Substitute dy/dt = 2(dx/dt) into the differentiated equation:
2y(2(dx/dt)) = 16(dx/dt).
5. Simplify and cancel out dx/dt (assuming it is not zero): 4y = 16.
6. Solve for y: y = 4.
7. Substitute y = 4 back into the original curve equation y² = 16x: 16 = 16x.
8. Solve for x: x = 1.
Therefore, the point on the curve is (1, 4).
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: