Question
Show that the perpendiculars drawn from the origin to tangent planes to the cone x2 y2 + 5z² + 4xy = 0 lie on the cone x2 y2 + z² + 4xy = 0.
Answer :
Word Count : 554
To solve this problem, we are given the equation of a cone: \[ x^2 y^2 + 5z^2 + 4xy = 0 \] and asked to show that the perpendiculars drawn from the origin to the tangent planes of this cone lie on the cone: \[ x^2 y^2 + z^2 + 4xy = 0 \] ### Step 1: Find the equation of the tangent plane For a surface defined by \( F(x, y, z) = 0 \), the equation of the tangent plane at a point \((x_0, y_0, z_0)\) is given by: \[ F_x(x_0, y_0, z_0)(x - x_0) + F_y(x_0, y_0, z_0)(y - y_0) + F_z(x_0, y_0, z_0)(z - z_0) = _____ __________ ___ ___ ____.
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To solve this problem, we are given the equation of a cone: \[ x^2 y^2 + 5z^2 + 4xy = 0 \] and asked to show that the perpendiculars drawn from the origin to the tangent planes of this cone lie on the cone: \[ x^2 y^2 + z^2 + 4xy = 0 \] ### Step 1: Find the equation of the tangent plane For a surface defined by \( F(x, y, z) = 0 \), the equation of the tangent plane at a point \((x_0, y_0, z_0)\) is given by: \[ F_x(x_0, y_0, z_0)(x - x_0) + F_y(x_0, y_0, z_0)(y - y_0) + F_z(x_0, y_0, z_0)(z - z_0) = _____ __________ ___ ___ ____.
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