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

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

\[ \Rightarrow 2y\frac{dy}{dx} = 6x\]

\[ \Rightarrow \frac{dy}{dx} = \frac{6x}{2y}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{3x}{y}\]

Let tangent at (h, k) pass through 

\[\left( \frac{4}{3}, 0 \right)\] .

Since, (h, k) lies on (i), we get

\[3 h^2 - k^2 = 8 . . . (ii)\]

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

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

\[(y - k) = \frac{3h}{k}(x - h) . . . (iii)\]

Since, the tangent passess through 

\[\left( \frac{4}{3}, 0 \right)\] .

\[\therefore (0 - k) = \frac{3h}{k}\left( \frac{4}{3} - h \right)\]

\[ \Rightarrow - k = \frac{4h}{k} - \frac{3 h^2}{k}\]

\[ \Rightarrow - k^2 = 4h - 3 h^2\]

\[\Rightarrow 8 - 3 h^2 = 4h - 3 h^2 \left[ \text { From } \left( ii \right) \right]\]

\[ \Rightarrow 8 = 4h\]

\[ \Rightarrow h = 2\]

Using (ii), we get

\[12 - k^2 = 8\]
\[ \Rightarrow k^2 = 4\]
\[ \Rightarrow k = \pm 2\]

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

\[(y - 2) = \frac{6}{2}(x - 2), \text { or }, 3x - y - 4 = 0, \text { and }\]
\[(y + 2) = \frac{6}{- 2}(x - 2), \text { or }, 3x + y - 4 = 0\]