Let the normal at a point P on the curve
y² – 3x² + y + 10 = 0 intersect the y-axis at ( 0,3/2 ) .
If m is the slope of the tangent at P to the curve, then |m| is equal to

Solution

Maths sir · 4 steps

Write the normal's slope two ways, once from the derivative and once from the given point, and equate them.

1. Slope of tangent. Differentiate implicitly: 2y y' - 6x + y' = 0, so y' = 6x/(2y+1). At P(h, k) the tangent slope is m = 6h/(2k+1), so the normal slope is -(2k+1)/6h.
2. Normal through the point. The normal joins P(h, k) to (0, 3/2), so its slope is also (k - 3/2)/h = (2k-3)/2h. Equate the two expressions for the normal slope.

   (2k-3)/2h = -(2k+1)/6h ⇒ 3(2k-3) = -(2k+1) ⇒ k = 1

3. Find h. P lies on the curve, so 1 - 3h² + 1 + 10 = 0, which gives h² = 4 and h = ± 2.
4. So. m = 6h/(2k+1) = (± 12)/3 = ± 4, so |m| = 4.

The common slip: 0.25. Reporting the slope of the normal instead of the tangent gives 0.25, not 4. The question asks for the tangent, so take m = 6h/(2k+1).

The idea to keep. When a normal or tangent passes through a given point, take the point of contact as (h, k) and write its slope two ways: from y' and from the two points. Then use the curve's equation.

Answer from the NTA final answer key. Solution by PrepWiser.