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What is the foot of the normal if the straight line x + y + 7 = 0 is normal to the hyperbola 3x2 – 4y2 = 12 whose normal is at the point (x1, y1)?

What is the foot of the normal if the straight line x + y + 7 = 0 is normal to the hyperbola 3x2 – 4y2 = 12 whose normal is at the point (x1, y1)? Correct Answer (-4, -3)

Equation of the given hyperbola is, 3x2 – 4y2 = 12 ……….(1) Differentiating both sides of (1) with respect to y we get, 3*2x(dy/dx) – 4*(2y) = 0 Or dx/dy = 4y/3x Therefore, the equation of the normal to the hyperbola (1) at the point (x1, y1) on it is, y – y1 = -(x1, y1) (x – x1) = -4y1/3x1(x – x1) Or 3x1y + 4y1x – 7x1y1 = 0 Now, if possible, let us assume that the straight line x + y + 7 = 0 ………..(2) This line is normal to the hyperbola (1) at the point (x1, y1). Then, the equation (2) and (3) must be identical. Hence, we have, 3x1/1 = 4y1/1 = -7x1y1/7 So, x1 = -4 and y1 = -3 Now, 3x12 – 4y12 = 3(-4)2 – 4(-3)2 = 12 This shows the point (-4, -3) lies on the hyperbola (1). So, it’s the normal to the hyperbola. Thus, it is evident that the straight line (3) is normal to the hyperbola (1); the co-ordinate foot is (-4, -3).

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