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Find the shortest distance from the line x + y = 4 and the parabola ${y}^2 + 4x + 4y = 0$

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Last updated date: 20th Jun 2024
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Answer
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Hint: Let us assume we have a point (a,b) on the given curve, and RP is the line from where we have to find the shortest distance (Refer the fig). Let’s find out the values of a and b, and with the help of distance formula, we’ll get the answer. 

Complete answer:

Now suppose PQ is a tangent to the curve at (a,b) now if this tangent and the line have the same slope then only the point (a,b) will have the shortest distance from the line now since the point lies on the curve.

Hence, that eventually is the shortest distance between the curve and the line.

Slope at (a,b) of the curve should be equal to the slope of the line.

 

Differentiating the curve ${y}^2 + 4x + 4y = 0$ with respect to x,

  $2y\dfrac{{dy}}{{dx}} + 4 + 4\dfrac{{dy}}{{dx}} = 0{\text{ so }}\dfrac{{dy}}{{dx}} = \dfrac{{ - 4}}{{\left( {2y + 4} \right)}}..............\left\{ 1 \right\}$

Differentiating the line with respect to x,

  ${\text{1 + }}\dfrac{{dy}}{{dx}} = 0{\text{ so }}\dfrac{{dy}}{{dx}} = - 1{\text{ }}.................................\left\{ 2 \right\}$

Eq1 = Eq2 

$\dfrac{{ - 4}}{{2y + 4}} = - 1{\text{ so y = 0 or b = 0 }}$

Now, (a,b) lies on parabola hence,

${{b}^2} + 4a + 4b = 0$ put b = 0

we get a = 0

 

So now we simply has to find the shortest distance of the point

(0,0) to the line x + y = 4

 

Formulae $\dfrac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$

  $\dfrac{{\left| {0 + 0 - 4} \right|}}{{\sqrt {{1^2} + {1^2}} }}{\text{ = }}\dfrac{4}{{\sqrt 2 }}$

 

Note - One must remember, the shortest distance b/w any two curves is always the specific line that is simultaneously perpendicular to both the curves.