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# Distance

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) $$\frac{1}{\sqrt{17}}$$

Note by Tanishq Varshney
2 years, 4 months ago

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$$4x^2+y^2+4x-4y+5=0$$

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. · 2 years, 4 months ago

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 · 2 years, 4 months ago

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

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

i dont have its answer, plz post ur method · 2 years, 4 months ago

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