Given, equation of cuve y2−2x3−4y+8=0. ⇒2yy′−6x2−4y′=0⇒y′=‌
3x2
y−2
Since, p(x,y) is a point on this curve. ∴‌‌‌
y−2
x−1
=‌
3x2
y−2
(for tangent lines passing through (1,2) ) ‌⇒‌‌(y−2)2=3x2(x−1) ‌∴‌‌y2−2x3−4y+8=0 ‌⇒‌‌(y−2)2−2x3+4=0 ‌⇒‌‌3x2(x−1)−2x3+4=0 ‌⇒‌‌3x3−3x2−2x3+4=0 ‌⇒‌‌x3−3x2+4=0 ‌⇒‌‌(x+1)(x−2)=0 ‌∴‌‌x=−1,2 For x=−1, there is no real value of y. For x=2;y=2±2√3 ∴‌‌y′=‌
3x2
y−2
=±2√3 Thus, tangent lines drawn from the point (1,2) is y=2±2√3(x−1). ∴ Two tangents are possible.