
Yefim S. answered 09/15/23
Math Tutor with Experience
2x/a2 + 2yy'/b2 = 0; y' = - xb2/(ya2) = m; y = - xb2/(ma2);
x2/a2 + x2b2/(m2a4) = 1; x2/a2(1 + b2/(m2a2)) =1 ; x = ± ma2/√(m2a2 + b2); y = - + b2/√(m2a2 + b2)
2 tangent lines: y = - b2/√(m2a2 + b2) + m(x - ma2/√(m2a2 + b2));
y = b2/√(m2a2 + b2) + m(x + ma2/√(m2a2 + b2));

Dayv O.
and, the denominator square root in the constant can be made=1 by the numerator. y1=mx-sqrt(b^2+(m^2)a^2),,,,, y2=mx+sqrt(b^2+(m^2)a^2)09/15/23