Let the eccentricity root of the left focus of the ellipse be 3 / 3, and the length of the line passing through point F and perpendicular to x cut by the ellipse be 4 root 3 / 3 1 elliptic equation Let AB be the left and right vertices of the ellipse,

Let the eccentricity root of the left focus of the ellipse be 3 / 3, and the length of the line passing through point F and perpendicular to x cut by the ellipse be 4 root 3 / 3 1 elliptic equation Let AB be the left and right vertices of the ellipse,




Point P is a point on the ellipse (x ^ 2 / 25) + (y ^ 2 / 16) = 1, m and N are the focal points, ∠ MPN = 60 °, and the area of △ MPN is calculated


The ellipse is: x ^ 2 / 5 ^ 2 + y ^ 2 / 4 ^ 2 = 1, so, a = 5, B = 4, so, C ^ 2 = a ^ 2-B ^ 2 = 25-16 = 9, so, C = 3, let PM = x, PN = y, then PM + PN = 2A = 10, that is, x + y = 10 (1) The cosine theorem is: Mn ^ 2 = PM ^ 2 + PN ^ 2-2 * PM * PN * cos ∠ MPN = x ^ 2 + y ^ 2-2xyc



If the triangle MPN is isosceles triangle, then the eccentricity of the ellipse is?
The answer is not root 2-1


A simple drawing shows that MPN is a right triangle, so it can only be PN = nm, PN is half of the path length, B ^ 2 / a Mn = 2C e ^ 2 + 2e-1 = 0. As a result, it's not convenient to type. Remember to round the one greater than 1



In the triangle MPN with the circumference of 48, ∠ MPN = 90 °, Tan ∠ PMN = 3 / 4, take M, n
In a triangle MPN with a circumference of 48, ∠ MPN = 90 ° and Tan ∠ PMN = 3 / 4, the hyperbolic equation with m, n as the focus and passing through point P is obtained


The Pythagorean theorem shows that Mn = 5K  perimeter is 48  3K + 4K + 5K = 48K = 4  Mn = 20 | pm-pn | = k = 4, assuming Mn as the x-axis and the midpoint o as the origin of rectangular coordinate system, then 2A = 4, 2C = 20  a = 2, C = 10B = √ (C  178; - a  178;) = 4  6  double