Question

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

Answer 1

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

Differentiate the expression wrt $x$

$\Rightarrow 6x-2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=3\left(\frac{x}{y}\right)$

Let the tangent passes through $(x_1,y_1)$ and it is given that it also passes through $(\frac{4}{3},0)$

$\therefore$ Slope $=\frac{y_1-0}{(x_1-\frac{y}{3})}=\frac{y_1}{(x_1-\frac{y}{3})}$

Equ $\left(1\right)$ can be written as ${\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=3\left(\frac{x_1}{y_1}\right)$

Equating Eq $\left(1\right)$ & $\left(2\right)$

$3\left(\frac{x_1}{y_1}\right)=\frac{y_1}{(x_1-\frac{4}{3})}\Rightarrow 3x_1(x_1-\frac{4}{3})=y_1^2\Rightarrow 3x_1^2-4x_1=y_1^2$ ....... $\left(1\right)$

Since $(x_1,y_1)$ is the contact point $(x_1,y_1)$ also satisfies the equation of curve.

$\Rightarrow 3x_1^2-y_1^2=8$ ..... $\left(2\right)$

Solving eq. $\left(1\right)$ and $\left(2\right)$

$\Rightarrow 4x_1=8\Rightarrow x_1=2y_1^2=3x_1^2-8\Rightarrow 3(2{)}^2-8=4y_1=\pm 2$

Eqn of tangents =$(y\pm 2)=3\left(\frac{2}{\pm 2}\right)(x-2)$

$\Rightarrow (y-2)=3(x-2)\Rightarrow y=3x-4$ and

$(y+2)=3(-1)(x-2)\Rightarrow y=-3x+4$