**Question:**

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

**Solution:**

assume point $(a, b)$ which lies on the given curve

finding the slope of the tangent by differentiating the curve

$6 x-2 y \frac{d y}{d x}=0$

$\frac{d y}{d x}=\frac{3 x}{y}$

m(tangent) at $(a, b)$ is $\frac{3 a}{b}$

Since this tangent passes through $\left(\frac{4}{3}, 0\right)$, its slope can also be written as

$\frac{b-0}{a-\frac{4}{3}}$

Equating both the slopes as they are of the same tangent

$\frac{b}{a-\frac{4}{3}}=\frac{3 a}{b}$

$b^{2}=3 a^{2}-4 a \ldots(i)$

Since points $(a, b)$ lies on this curve

$3 a^{2}-b^{2}=8$

Solving (i) and (ii) we get

$3 a^{2}-8=3 a^{2}-4 a$

$a=2$

$b=2$ or $-2$

therefore points are $(2,2)$ or $(2,-2)$

equation of tangent is given by $y-y_{1}=m(\operatorname{tangent})\left(x-x_{1}\right)$

$y-2=3(x-2)$

Or

$y+2=-3(x-3)$