Problem Presentation and Solution

The problem at hand is to find the value of ( p ) such that the tangent to the curve ( y px^3 ) at the point ( x 2 ) passes through the point ( (1, -10) ).

Let's go through the steps systematically to solve this problem:

Step 1: Determine the Point on the Curve

The point on the curve where ( x 2 ) is calculated as:

( y p cdot 2^3 8p )

Thus, the point on the curve is ( (2, 8p) ).

Step 2: Find the Derivative to Determine the Slope of the Tangent

The derivative of ( y px^3 ) with respect to ( x ) is:

( frac{dy}{dx} 3px^2 )

At ( x 2 ), the slope of the tangent is:

( frac{dy}{dx}bigg|_{x2} 3p cdot 2^2 12p )

Step 3: Write the Equation of the Tangent Line

The equation of the tangent line at the point ( (2, 8p) ) with slope ( 12p ) is given by:

( y - 8p 12p(x - 2) )

Simplifying this, we get:

( y - 8p 12px - 24p )

( y 12px - 16p )

Step 4: Substitute the Point (1, -10) into the Tangent Line Equation

Substituting ( x 1 ) and ( y -10 ) into the tangent line equation, we get:

( -10 12p(1) - 16p )

( -10 12p - 16p )

( -10 -4p )

( p frac{10}{4} frac{5}{2} )

Conclusion

Thus, the value of ( p ) is ( frac{5}{2} ).

Graphical Verification

For verification, let's verify the solution graphically:

The curve is given by ( y frac{5}{2}x^3 ). At ( x 2 ), we have ( y frac{5}{2} cdot 8 20 ). Thus, the point on the curve is ( (2, 20) ).

The slope of the tangent line at ( x 2 ) is:

( frac{dy}{dx}bigg|_{x2} 12 cdot frac{5}{2} 30 )

The equation of the tangent line at the point ( (2, 20) ) is:

( y 30(x - 2) 20 )

( y 3 - 60 20 )

( y 3 - 40 )

This line indeed passes through the point ( (1, -10) ), since:

( -10 30 cdot 1 - 40 )

( -10 -10 )

Hence, our solution is correct.