Question

Find the equation of the tangents to the curve 3x² – y² = 8, which passes through the point (4/3, 0).

Answer 1

We have,

3x² – y² = 8 ...(i)

Differentiating both sides w.r.t x, we get

$$\begin{gathered} 6x-2y\frac{dy}{dx}=0 \\ \Rightarrow 2y\frac{dy}{dx}=6x \\ \Rightarrow \frac{dy}{dx}=\frac{6x}{2y} \\ \Rightarrow \frac{dy}{dx}=\frac{3x}{y} \end{gathered}$$

Let tangent at (h, k) pass through $\left(\frac{4}{3},0\right)$.

Since, (h, k) lies on (i), we get

$3h^2-k^2=8...\left(\mathrm{ii}\right)$

Slope of tangent at (h, k) = $\frac{3h}{k}$

The equation of tangent at (h, k) is given by,

$(y-k)=\frac{3h}{k}(x-h)...\left(\mathrm{iii}\right)$

Since, the tangent passess through $\left(\frac{4}{3},0\right)$.

$$\begin{gathered} \therefore (0-k)=\frac{3h}{k}\left(\frac{4}{3}-h\right) \\ \Rightarrow -k=\frac{4h}{k}-\frac{3h^2}{k} \\ \Rightarrow -k^2=4h-3h^2 \end{gathered}$$

$$\begin{gathered} \Rightarrow 8-3h^2=4h-3h^2\left[\mathrm{From}\left(\mathrm{ii}\right)\right] \\ \Rightarrow 8=4h \\ \Rightarrow h=2 \end{gathered}$$

Using (ii), we get

$$\begin{gathered} 12-k^2=8 \\ \Rightarrow k^2=4 \\ \Rightarrow k=\pm 2 \end{gathered}$$

So, the points on curve (i) at which tangents pass through $\left(\frac{4}{3},0\right)$ are $\left(2,\pm 2\right)$.

Now, from (iii), the equation of tangents are
$$\begin{gathered} (y-2)=\frac{6}{2}(x-2),\mathrm{or},3x-y-4=0,\mathrm{and} \\ (y+2)=\frac{6}{-2}(x-2),\mathrm{or},3x+y-4=0 \end{gathered}$$