It is known that ⊙ o is a circle with the origin as the center and √ 2 as the radius. The point P is a point on the straight line y = - x + 6. Through P, make a tangent PQ of ⊙ o, and Q is the tangent point. Then the minimum value of the tangent length PQ is

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• 1. Make the tangent of circle (x - 6) ^ 2 + y ^ 2 = 4 through origin 0, then the tangent length is
• 2. Make the tangent of the circle (X-6) 2 + y2 = 4 through the origin O & nbsp;, and the length of the tangent is______ ．
• 3. Make all lines at a point a on the curve y = x * 2 (x is greater than or equal to 0) so that the area enclosed by the curve and X axis is 1 / 12. Try to find the tangent point a and tangent square
• 4. Find the special solution of the differential equation y '' '= e ^ (- x) satisfying the initial condition y | (x = 1) = y' | (x = 1) = y '' | (x = 1) = 0
• 5. The special solution of the differential equation y '= e ^ x + y satisfying the condition y (0) = 0 is
• 6. A special solution to the differential equation y '+ 2XY = e ^ (- x ^ 2) satisfying the initial condition y (0) = 2
• 7. The special solution of the differential equation y '' - y = e ^ x satisfying the condition y (0) = 0, y '(0) = 0 is
• 8. The solution of differential equation y '= 2x (y * Y-Y') satisfying initial condition y (0) = 1 is as follows
• 9. Given that the general formula of a differential equation is y = 2x + C, then the special solution of the differential equation under the initial condition x = O, y = 3 is?
• 10. Special solutions of differential equation y '- y = 2xe ^ (2x), y (0) = 1
• 11. Given that a curve is located on the right side of the Y-axis and passes through the point (1.1), the intercept of the tangent at any point on the curve on the y-axis is equal to the abscissa of the tangent point, the curve equation is obtained
• 12. Finding the normal equation The normal equation of y = x ^ 3-x + 5 at point m (0,5)
• 13. The normal equation of curve y = x ^ 2 + 1 at point (1,2) is?
• 14. The normal equation of a curve at a certain point Finding the normal equation of the curve y = e ^ - 2x at M (0,1) The answer is y '= - 2E ^ - 2x, so the tangent slope is - 2, so the normal slope is - 1 / (tangent slope) = 1 / 2 So the normal equation Y-1 = 1 / 2 (x-0) I don't know why "so normal slope is - 1 / (tangent slope) = 1 / 2" I don't know who can help me popularize the concept of normals and general methods
• 15. It is proved that the length of the line segment between two coordinate axes of the tangent at any point on the curve X ^ 2 / 3 + y ^ 2 / 3 = a ^ 2 / 3 (note a > 0 constant, 2 / 3 is power) is fixed
• 16. Find the curve of a point (1, - 1) so that the tangent line at any point on the curve is bisected by the tangent line
• 17. When a curve passes through a point (2,3), any line segment between two coordinate axes is bisected by tangent point
• 18. A curve passes through point (2,3). Any tangent line between its two axes is bisected by the tangent point. The equation of this line is obtained
• 19. A curve on the xoy plane passes through the point (2,3), and any line segment between the two coordinate axes is bisected by the tangent point According to the meaning of the topic, the intercept of the tangent of the curve at the point (x, y) on the two coordinate axes should be 2x and 2Y How do we get the intercept 2x and 2Y?
• 20. When a curve passes through a point (2,3), any line segment between two coordinate axes is bisected by the tangent point?