Question

The equation of tangent at (2, 3) on the curve y² = px³ + q is y = 4x – 7. Find the values of  ‘p’ and ‘q’.

Answer 1

Given: Equation of curve is y² = px³ + q

Differentiating both sides w.r.t. x, we get

`2y dy/dx = 3px^2`

⇒ `dy/dx = (3px^2)/(2y)`

∴ `(dy/dx)_((2"," 3)) = (3p xx 2^2)/(2 xx 3)`

or `m = (dy/dx)_((2","3))`

= `(12p)/6`

= 2p

Since, y = 4x – 7 is the tangent to the curve at point (2, 3).

So, on comparing with y = mx + c, we get

m = 4

Now, 2p = 4

⇒ `p = 4/2`

⇒ p = 2

Since, point (2, 3) lies on the curve,

∴ 3² = p × 2³ + q

⇒ 9 = 2 × 8 + q   ...[∵ p = 2]

⇒ 9 – 16 = q

⇒ q = –7

Hence, p = 2 and q = –7.