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Let points `A_(1), A_(2) and A_(3)` lie on the parabola `y^(2)=8x.` If `triangle A_(1)A_(2)A_(3)` is an equilateral triangle and normals at points `A_(1), A_(2) and A_(3)` on this parabola meet at the point (h, 0). Then the value of h I s

A

24

B

26

C

38

D

28

Text Solution

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To solve the problem, we need to find the value of \( h \) where the normals at points \( A_1, A_2, A_3 \) on the parabola \( y^2 = 8x \) meet at the point \( (h, 0) \). The points \( A_1, A_2, A_3 \) form an equilateral triangle. ### Step 1: Identify the points on the parabola The parabola \( y^2 = 8x \) can be rewritten in the standard form \( y^2 = 4Ax \) where \( A = 2 \). The parametric coordinates of points on the parabola can be expressed as: \[ A_i = (2t_i^2, 4t_i) \quad \text{for } i = 1, 2, 3 \] ### Step 2: Determine the angles for the equilateral triangle Since \( A_1, A_2, A_3 \) form an equilateral triangle, the angles between the normals at these points will be \( 60^\circ \). The angles can be represented in radians as \( \frac{\pi}{3} \). ### Step 3: Find the slopes of the normals The slope of the tangent to the parabola at point \( A_i \) is given by the derivative: \[ \frac{dy}{dx} = \frac{4}{2t_i} = \frac{2}{t_i} \] Thus, the slope of the normal at \( A_i \) is: \[ -\frac{t_i}{2} \] ### Step 4: Write the equations of the normals The equation of the normal at point \( A_i \) can be written as: \[ y - 4t_i = -\frac{t_i}{2}(x - 2t_i^2) \] Rearranging gives: \[ y = -\frac{t_i}{2}x + 4t_i + t_i^3 \] ### Step 5: Substitute \( (h, 0) \) into the normal equations Since all normals meet at \( (h, 0) \), we can substitute \( y = 0 \) into the normal equations: \[ 0 = -\frac{t_i}{2}h + 4t_i + 2t_i^3 \] This simplifies to: \[ \frac{t_i}{2}h = 4t_i + 2t_i^3 \] Dividing by \( t_i \) (assuming \( t_i \neq 0 \)): \[ \frac{h}{2} = 4 + 2t_i^2 \] Thus, \[ h = 8 + 4t_i^2 \] ### Step 6: Find the value of \( t_i \) Since the points are symmetric and form an equilateral triangle, we can assume \( t_1 = 2\sqrt{3} \) (as derived from the angle considerations). Therefore, substituting \( t_i = 2\sqrt{3} \) into the equation gives: \[ h = 8 + 4(2\sqrt{3})^2 \] Calculating \( (2\sqrt{3})^2 = 4 \cdot 3 = 12 \): \[ h = 8 + 4 \cdot 12 = 8 + 48 = 56 \] ### Final Answer Thus, the value of \( h \) is: \[ \boxed{56} \]
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