Question

Find the equation of tangent to the curve $y^2-7x-8y+14=0$ at the point $(2,0)$.

Answer 1

Differentiate wrt $x$

$\therefore 2y\frac{dy}{dx}-7-8\frac{dy}{dx}+14=0$

$\therefore \frac{dy}{dx}{|}_{(}{x}_1,y_1)=\frac{-7}{2y_1-8}$

Hence, slope at (2,0) is $\frac{-7}{4-8}=\frac{7}{4}$

$⟹y=\frac{7}{4}x+c$

substitute point to get

$c=-\frac{7}{2}$

$⟹$ equation of tangent is $y=\frac{7}{4}x-\frac{7}{2}$

$⟹7x-4y=14$