Find the equations of the tangents to the curve 3x²- y²= 8, which pass through the point (4/3.0).

Question

Find the equations of the tangents to the curve
3x2- y2= 8, which pass through the point (4/3
0)

Gursheen kaur. · Accepted Answer

The equation of the given curve is

$3 x^{2} - y^{2} = 8................. (i)$

Since (4/3,0) lies on (i). Therefore , the coordinates of the given point are (4/3 ,0).

Now,

$3 x^{2} - y^{2} = 8 D i f f e r e n t i a t i n g w . r . t' x', w e g e t, \Rightarrow 6 x - 2 y \frac{d y}{d x} = 0 \Rightarrow 6 x = 2 y \frac{d y}{d x} o r \frac{d y}{d x} = \frac{6 x}{2 y} = \frac{3 x}{y} {(\frac{d y}{d x})}_{(\frac{4}{3}, 0)} = \frac{3 \times \frac{4}{3}}{0} = 0$

The equation of the tangent at (4/3,0) is

$(y - 0) = {(\frac{d y}{d x})}_{(\frac{4}{3}, 0)} (x - \frac{4}{3}) \Rightarrow y = 0 (x - \frac{4}{3}) \Rightarrow y = 0$

Find the equations of the tangents to the curve 3x2- y2= 8, which pass through the point (4/3.0).

Find the equations of the tangents to the curve 3x²- y²= 8, which pass through the point (4/3.0).