Find the length of tangent of $y^{2} = 12 x$ at $(12, 12)$

12 \sqrt{5}

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14 \sqrt{5}

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16 \sqrt{5}

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None of the above

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Solution

The correct option is A $12 \sqrt{5}$
Given equation is $y^{2} = 12 x$ at $(12, 12)$
Length of the tangent is $=$ $y \frac{\sqrt{1 + m^{2}}}{m}$
$= 12 \frac{\sqrt{1 + {\frac{(1)}{(2)^{2}}}^{2}}}{\frac{1}{2}}$
$= 12 \sqrt{5}$

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Find the length of tangent of y2=12x at (12,12)

The correct option is A 12√5Given equation is y2=12x at (12,12)Length of the tangent is = y√1+m2m=12√1+(1)(2)2212=12√5

Find the length of tangent of $y^{2} = 12 x$ at $(12, 12)$

The correct option is A $12 \sqrt{5}$
Given equation is $y^{2} = 12 x$ at $(12, 12)$
Length of the tangent is $=$ $y \frac{\sqrt{1 + m^{2}}}{m}$
$= 12 \frac{\sqrt{1 + {\frac{(1)}{(2)^{2}}}^{2}}}{\frac{1}{2}}$
$= 12 \sqrt{5}$