Question

If $y = 4x - 5$ is a tangent to the curve $y^{2} = px^{3} + q$ at $(2, 3)$, then $p-q$ is

• A. -5
• B. 5
• C. 9
• D. -9

Hint:

Remember that the slope of the tangent to a curve at a given point is found by evaluating the derivative of the curve's equation at that point. If a line is tangent to a curve, their slopes at the point of tangency must be equal, and the point of tangency must satisfy both the curve's and the line's equations.

Answer 1

The Correct Option is A

Step 1: The given curve is $y^{2} = px^{3} + q$. Differentiate both sides with respect to $x$.\ \[ 2y \frac{dy}{dx} = 3px^{2} \] \[ \frac{dy}{dx} = \frac{3px^{2{2y} \]\
Step 2: The point of tangency is $(2, 3)$. Substitute $x=2$ and $y=3$ into the derivative to find the slope of the tangent at this point.\ \[ \left(\frac{dy}{dx}\right)_{(2,3)} = \frac{3p(2)^{2{2(3)} = \frac{3p(4)}{6} = \frac{12p}{6} = 2p \]\
Step 3: The equation of the tangent line is given as $y = 4x - 5$. The slope of this line is $4$.\
Step 4: Since $y = 4x - 5$ is tangent to the curve, the slope of the tangent at $(2,3)$ must be equal to the slope of the line.\ \[ 2p = 4 \] \[ p = 2 \]\
Step 5: The point $(2, 3)$ lies on the curve $y^{2} = px^{3} + q$. Substitute $x=2$, $y=3$, and $p=2$ into the curve's equation.\ \[ (3)^{2} = (2)(2)^{3} + q \] \[ 9 = 2(8) + q \] \[ 9 = 16 + q \] \[ q = 9 - 16 \] \[ q = -7 \]\
Step 6: Calculate $p-q$.\ \[ p-q = 2 - (-7) \] \[ p-q = 2 + 7 \] \[ p-q = 9 \]