Question

Find the equation of the tangent to the curve 3x ^2-y^2=8 which passes through the point (4/3,0)

Answer 1

"Let it be y=mx+c,

Then, let it touch the curve at a point (a,b).

3â2−b2=8, and also

b=ma+c

Substituting b into the first equation and resolving, we obtain,

(3−m2)a2−2mca−(c2+8)=0

Now because a tangent will intersect a 2nd degree curve such as this at only one point, the 2 roots of this quadratic in a will be equal, which means that the discriminant will vanish.

D=0

(−2mc)2=−4⋅(3−m2)(c2+8)

Simplifying this, we obtain,

8m2−3c2–24=0…(i)

Now, we know that our tangent line is passing through the point (43,0) too. So,

0=4m3+c

m=−3c4

Replacing m in (i), and solving the quadratic in c, we get,

c=±4 and thus m=∓3

∴y=∓3x±4

are our required tangents."