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:
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
Let the focal chord PQ of the parabola $ y^2 = 4x $ make an angle of $ 60^\circ $ with the positive x-axis, where P lies in the first quadrant. If the circle, whose one diameter is PS, $ S $ being the focus of the parabola, touches the y-axis at the point $ (0, \alpha) $, then $ 5\alpha^2 $ is equal to:
Rearrange the following parts to form a meaningful and grammatically correct sentence:
P. a healthy diet and regular exercise
Q. are important habits
R. that help maintain good physical and mental health
S. especially in today's busy world