Question

If a tangent drawn at the point \(P(h,k)\), where \(h,k\in \mathbb{Z}\), on the curve \[ y=2x^3+3x^2-4x-1 \] passes through the point \(Q(2,8)\), then \(PQ=\)

• A. \(\sqrt{65}\)
• B. \(5\)
• C. \(13\)
• D. \(\sqrt{85}\)

Hint:

When a tangent passes through a fixed external point, write the tangent equation at a general point \((h,f(h))\) and substitute the external point into it. This converts the problem into an algebraic equation in \(h\).

Answer 1

The Correct Option is A

Concept: If the tangent at a point \(P(h,k)\) on a curve passes through a fixed point \(Q\), then the coordinates of \(P\) must satisfy both:
• The point lies on the curve.
• The tangent equation passes through the given external point. We first find the tangent at a general point \(x=h\), then impose the condition that \(Q(2,8)\) lies on that tangent.

Step 1: Differentiate the curve. Given \[ y=2x^3+3x^2-4x-1. \] Differentiating, \[ \frac{dy}{dx} = 6x^2+6x-4. \] Hence the slope of the tangent at \(x=h\) is \[ m=6h^2+6h-4. \]

Step 2: Coordinates of the point of tangency. Since \(P(h,k)\) lies on the curve, \[ k=2h^3+3h^2-4h-1. \] Therefore, \[ P=\Bigl(h,\;2h^3+3h^2-4h-1\Bigr). \]

Step 3: Equation of the tangent. The tangent at \(P\) is \[ y-k = (6h^2+6h-4)(x-h). \] Since \(Q(2,8)\) lies on this tangent, \[ 8-k = (6h^2+6h-4)(2-h). \] Substituting \[ k=2h^3+3h^2-4h-1, \] we get \[ 9-2h^3-3h^2+4h = (6h^2+6h-4)(2-h). \] Expanding the right-hand side, \[ = 12h^2+12h-8 -6h^3-6h^2+4h. \] \[ = -6h^3+6h^2+16h-8. \] Hence, \[ 9-2h^3-3h^2+4h = -6h^3+6h^2+16h-8. \] Simplifying, \[ 4h^3-9h^2-12h+17=0. \]

Step 4: Find the integral root. Since \(h\in\mathbb Z\), test integral values. Substituting \(h=1\), \[ 4-9-12+17=0. \] Therefore, \[ h=1. \] Hence, \[ k = 2(1)^3+3(1)^2-4(1)-1 = 0. \] Thus, \[ P=(1,0). \]

Step 5: Calculate \(PQ\). Using the distance formula, \[ PQ = \sqrt{(2-1)^2+(8-0)^2}. \] \[ = \sqrt{1+64}. \] \[ = \sqrt{65}. \] Hence, \[ \boxed{\sqrt{65}}. \]