Write out the equations of two curves passing through point a (- 5,3) and tangent to the curve xy = 1 The title is changed to write out the equations of two straight lines passing through point a (- 5,3) and tangent to the curve xy = 1

Write out the equations of two curves passing through point a (- 5,3) and tangent to the curve xy = 1 The title is changed to write out the equations of two straight lines passing through point a (- 5,3) and tangent to the curve xy = 1

y=1/x
y'=-1/x^2
Let the tangent point be (a, 1 / a), then
(1/a-3)/(a+5)=-1/a^2
3a^2-2a-5=0
A = - 1 or 5 / 3
So the tangent is x + y + 2 = 0 or 9x + 25y-30 = 0

The vertex a of the regular triangle ABC is the right vertex of the hyperbola x ^ 2 + ay ^ 2 = 1, and the vertices B and C are on the right branch of the hyperbola. What is the value range of a

A (1,0) because the triangle ABC is an equilateral triangle, so AB = BC suppose B (C, radical [(1-C ^ 2) / a]), C (C, - radical [(1-C ^ 2) / a]) AB = radical [(1-C ^ 2) / 2 + (C-1) ^ 2], BC = 2 * radical [(1-C ^ 2) / a] to get the equation about C and a, (1 + 3 / a) * C ^ 2-2c + 1-3 / a = 0

It is known that the right vertex of hyperbola x ^ 2-my ^ 2 = 1 is a, and B.C is two points on its right branch. If the triangle ABC is an equilateral triangle, then M belongs to?
Why is not the slope of the asymptote greater than 30?

Have you ever thought that if you take any point m (except a) on the right branch and connect am, then the straight line am must have an intersection with the right asymptote (which can be verified by a straight line and a hyperbola). This does not mean that the inclination angle of am must be greater than that of the asymptote (M is equal when a is close to infinity). Therefore, if the inclination angle of an equilateral triangle is 30 degrees, then the inclination angle of the asymptote should be less than 30 degrees
That's good