Question

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

Answer 1

Find equation of tangent to

curve $3x^2-y^2=8$ passing

via point $(\frac{4}{3},0)$

curve $\to 3x^2-y^2=8$

differentiating on both side

with respect to 'x'

$3\times 2x-2\frac{dy}{dx}=0$

$-2\frac{dy}{dx}=-6x$

$\frac{dy}{dx}=\frac{6x}{2}=3x$

For equation of tangent

point $\to (\frac{4}{3},0)$

slope $\to 3x=3\times \frac{4}{3}=4$

$\frac{y-y_0}{x-x_0}=m\Rightarrow \frac{y-0}{x-\left(\frac{4}{3}\right)}=4$

$y=4(x-\frac{4}{3})$

$y=4x-\frac{16}{3}$