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

$3x−y−4=0$ and $3x+y−4=0$

The correct answer is Option (1) → $3x−y−4=0$ and $3x+y−4=0$

The given curve is $3x^2 - y^2 = 8$   ...(i)

Let $P(x_1, y_1)$ be a point on the curve (i) the tangent at which passes through the point $(\frac{4}{3},0)$.

As $P(x_1,y_1)$ lies on the curve $3{x_1}^2-{y_1}^2=8$

Differentiating (i) w.r.t. x, we get

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

$⇒\frac{dy}{dx}=\frac{3x}{y}$

∴ Slope of tangent at $P(x_1,y_1) =\frac{3x_1}{y_1}$

∴ The equation of tangent to the curve (i) at $P(x_1,y_1)$ is

$y-y_1 =\frac{3x_1}{y_1} (x-x_1)$.

Since it passes through the point $(\frac{4}{3},0)$, we get

$0-y_1=\frac{3x_1}{y_1}(\frac{4}{3}-x_1)$

$⇒-{y_1}^2=4x_1-3{x_1}^2$

$⇒3{x_1}^2-{y_1}^2=4x_1⇒4x1 = 8$  (using (ii))

$⇒x_1 = 2$.

Substituting this value of $x_1$ in equation (ii), we get

$3 × 2^2-{y_1}^2=8$

Thus, we get the points $P(2, ±2)$ on the curve (i), the tangent at which passes through the point $(\frac{4}{3},0)$.

∴ Equations of tangents to the curve are

$y-2=\frac{3 × 2}{2}(x-2)$ and $y+2=\frac{3 × 2}{-2}(x-2)$

$⇒y-2=3(x-2)$ and $y+2=-3(x-2)$

$⇒3x-y-4-0$ and $3x + y - 4 = 0$.

Hence, the equations of required tangents are $3x-y-4=0$ and $3x+y-4=0$.