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Find a curve from a geometric property of its tangent or normal

This type of question, asked 4 times: 2021 Paper 26, 14 Jun 2022, 6 Apr 2023, 30 Jan 2024. All 4 questions below, oldest first.

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  1. Question 1

    2021 Paper 26

    88
    Let the normals at all the points on a given curve pass through a fixed point (a,b)(a,b). If the curve passes through (3,−3)(3, - 3) and (4,−22)(4, - 2\sqrt{2}), and given that a−22 b=3a- 2\sqrt{2}\text{ }b = 3, then (a2+b2+ab)\left( a^{2}+b^{2}+ ab \right) is equal to

  2. Question 2

    14 Jun 2022

    158
    A particle is moving in the xy-plane along a curve C passing through the point (3,3)(3,3). The tangent to the curve CC at the point PP meets the XX-axis at QQ. If the yy-axis bisects the segment PQPQ, then CC is a parabola with

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  3. Question 3

    6 Apr 2023

    178
    Let a curve y=f(x),x∈(0,∞)y = f(x),x \in (0,\infty) pass through the points P(1,32)P\left( 1,\frac{3}{2} \right) and Q(a,12)Q\left( a,\frac{1}{2} \right). If the tangent at any point R(b,f(b))R(b,f(b)) to the given curve cuts the yy-axis at the point S(0,c)S(0,c) such that bc=3bc = 3, then (PQ)2(PQ)^{2} is equal to

  4. Question 4

    30 Jan 2024

    171
    Let Y=Y(X)Y = Y(X) be a curve lying in the first quadrant such that the area enclosed by the line Y−y=Y′(x)(X−x)Y - y =Y^{'}(x)(X - x) and the co-ordinate axes, where ( x,yx,y ) is any point on the curve, is always −y22Y′(x)+1,Y′(x)≠0\frac{-y^{2}}{2Y^{'}(x)}+ 1,Y^{'}(x) \neq 0. If Y(1)=1Y(1) = 1, then 12Y(2)12Y(2) equals