The minimum distance of \(4x^{2}+y^{2}+4x-4y+5=0\) from the line \(-4x+y=3\) is??

options 1)2 \(\quad\) 2)0 \(\quad\) 3)1 \(\quad\) 4) \(\frac{1}{\sqrt{17}}\)

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## Comments

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TopNewest\(4x^2+y^2+4x-4y+5=0\)

Rewriting it, we get:

\(4x^2+4x+1+y^2-4y+4=0\).

\((2x+1)^2+(y-2)^2=0\).

Since, \(LHS\) is always non-negative, both the square terms are equal to \(0\).

This gives us \(x=\dfrac{-1}{2}\) and \(y=2\).

Now, we have to find the minimum distance between \((\dfrac{-1}{2},2)\) and the line \(-4x+y=3\) which in other words is our perpendicular distance between the point and the line.

Slope of the line \(-4x+y=3\) is equal to \(4\) and so the slope of the line perpendicular to \(-4x+y=3\) will be \(\dfrac{-1}{4}\) with \((\dfrac{-1}{2},2)\) lying on it because the product of the slopes of the \(2\) perpendicular lines is equal to \(-1\).

Using the point-slope formula, we can get the equation of the second line as \(2x+8y=15\).

Now, we have to find the intersection point of both the lines and then take the distance of the intersection point from our original point \((\dfrac{-1}{2},2)\).

Solving our pair of equations, we get \(x=\dfrac{-9}{34}\) and \(y=\dfrac{33}{17}\).

Now, applying the Distance Formula between the points \((\dfrac{-1}{2},2)\) and \((\dfrac{-9}{34},\dfrac{33}{17})\), we get

\(\dfrac{1}{\sqrt{17}}\)as our answer.Log in to reply

Answer comes out to be 4. The above curve reduces to. (2x +1 )^2 + ( y-2)^2 =0.which implies x=-1/2 and y=2 whose perpendicular distance from given line turns out to be 1/(17)^0.5

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\(4x^{2}+y^{2}+4x-4y+5=0=(2x+1)^2 + (y-2)^2 \Rightarrow (x=\frac{-1}{2}\) , \( y=2)\)

Now find the perpendicular distance from the point P \( (\frac{-1}{2} , 2)\) to the curve \(-4x+y=3\) by using \( \frac{ax_1 + by_1 + c_1}{\sqrt{a^2+b^2}} = \dfrac{1}{\sqrt{17}}\)

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i dont have its answer, plz post ur method

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@Yash Singhal @rachit parikh @Saurabh Patil @Nishant Rai

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