Find the equation to that tangent to the parabola $y^{2} = 7 x$ which is parallel to the straight line 4y - x + 3 = 0. Find also its point of contact.

Open in App

Solution

$4 y - x + 3 = 0$

Slope of line $= m = - \frac{a}{b} = - \frac{- 1}{4} = \frac{1}{4}$

For $y^{2} = 7 x$

$4 a = 7 \Rightarrow a = \frac{7}{4}$

Equation of tangent is

$y = m x + \frac{a}{m} y = \frac{1}{4} x + \frac{\frac{7}{4}}{\frac{1}{4}} y = \frac{1}{4} x + 7 4 y = x + 28$

$\Rightarrow x = 4 y - 28$

Substituting $x$ in equation of parabola

$y^{2} = 7 (4 y - 28) y^{2} - 28 y + 196 = 0 {(y - 14)}^{2} = 0 \Rightarrow y = 14 x = 4 y - 28 \Rightarrow x = 4 (14) - 28 = 28$

So, the point of contact is $(28, 14)$ .

Suggest Corrections

Similar questions

y = x^{2} - 7 x + 3

5 x + y - 3 = 0.

x^{2} + y^{2} - 2 x + 4 y = 0

y^{2} = 16 x

2 x - y + 5 = 0

y = x^{2} - 3 x + 2

5 x - 2 y + 10 = 0

y^{2} = 16 x

2 x - y + 5 = 0.

Join BYJU'S Learning Program