Question

find the equation of yhe tangent to the parabola y^ = 12x from the point (2,5)
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Answer 1

Question :-

→ Find the equation of the tangent to the parabola y² = 12x from the point (2,5)

Answer :-

$\mapsto \boxed{\sf \: 5y - 6x = 13} \:$

To Find :-

→ Equation of tangent .

Solution :-

If we have any parabola y² = ax ,

firstly we have to find gradient (slope ) at given point by differentiate to the equation of parabola .

We have ,

→ y² = 12x

differentiate with respect to "x"

$\mapsto \sf \: 2y \: \frac{dy}{dx} = 12 \\ \: \\ \mapsto \sf\: (slope) \: \frac{dy}{dx} = \frac{6}{y} \\ \\ \sf \: slope \: at \: (2,5) \\ \\ \mapsto \boxed{ \sf \: \frac{dy}{dx} \bigg |_{(2,5)} \: = \frac{6}{5} }$

$</p><p> \sf If \: we \: have \: slope (m) \: and \: passing \: \\ \sf \: points \: (x_1 ,y_1) \: \\ \sf so \: equation \: of \: tangent \: is \: - \\ \\ \to \boxed{ \sf \: y - y_1 = m(x - x_1)}$

Hence , equation of tanget passes through (2,5) is -

$\mapsto \sf \: y - 5 = \frac{6}{5} (x - 2) \\ \\ \mapsto \sf \: 5y - 25 = 6x - 12 \\ \\ \mapsto \boxed{\sf \: 5y - 6x = 13}$