Question

the tangents drawn from the ends of a latus rectum of y^2=12x at​

Answer 1

SOLUTION

TO DETERMINE

The tangents drawn from the ends of a latus rectum of y² = 12x

EVALUATION

Here the given equation of the parabola is

y² = 12x - - - - - (1)

Comparing with the general equation y² = 4ax we get

4a = 12

⇒ a = 3

So the ends of the latus rectum are (3, 6) & (3, - 6)

Differentiating both sides with respect to x we get

$\displaystyle \sf{2y \frac{dy}{dx} = 12 }$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = \frac{12}{2y} }$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = \frac{6}{y} }$

$\displaystyle \sf{ \frac{dy}{dx} \bigg|_{(3,6)} = \frac{6}{6} = 1}$

$\displaystyle \sf{ \frac{dy}{dx} \bigg|_{(3, - 6)} = \frac{6}{ - 6} = - 1}$

The required equation of the tangent at (3,6) is

$\displaystyle \sf{(y - 6) = 1(x - 3) }$

$\displaystyle \sf{ \implies \: x - y + 3 = 0 }$

The required equation of the tangent at (3, - 6) is

$\displaystyle \sf{(y + 6) = - 1(x - 3) }$

$\displaystyle \sf{ \implies \: x + y + 9 = 0 }$