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

#Geometry

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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Comments

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

rachit parikh - 6 years, 1 month ago

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

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.

Yash Singhal - 6 years, 1 month ago

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

Tanishq Varshney - 6 years, 1 month ago

i dont have its answer, plz post ur method

Tanishq Varshney - 6 years, 1 month 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 $\frac{ax_1 + by_1 + c_1}{\sqrt{a^2+b^2}} = \dfrac{1}{\sqrt{17}}$

Nishant Rai - 6 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$