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).
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
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: