Question:

The focus of the curve $y^2 + 4x - 6y + 13 = 0$ is

Updated On: May 16, 2024
  • (2, 3)
  • (-2, 3)
  • (2, -3)
  • (-2, -3)
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The Correct Option is B

Solution and Explanation

The given equation of curve is :
$y^2 + 4x - 6y + 13 = 0$
which can be written as : $y^2 - 6y + 9 + 4x + 4 = 0$
$\Rightarrow \left(y^{2} - 6y + 9\right) = - 4\left(x +1\right)$
$\Rightarrow \left(y-3\right)^{2} = -4\left(x+ 1\right)$
Put $Y = y - 3$ and $X = x + 1$
On comparing $Y^{2} = 4aX$
Length of focus from vertex, $a = - 1$
At focus $X = a$ and $Y = 0 \Rightarrow x + 1 = - 1$
$\Rightarrow x = - 2$
$\therefore y-3 = 0 \Rightarrow y=3$
$\therefore$ Focus is $\left(- 2, 3\right).$
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Concepts Used:

Equation of a Line in Space

In a plane, the equation of a line is given by the popular equation y = m x + C. Let's look at how the equation of a line is written in vector form and Cartesian form.

Vector Equation

Consider a line that passes through a given point, say ‘A’, and the line is parallel to a given vector '\(\vec{b}\)‘. Here, the line ’l' is given to pass through ‘A’, whose position vector is given by '\(\vec{a}\)‘.  Now, consider another arbitrary point ’P' on the given line, where the position vector of 'P' is given by '\(\vec{r}\)'.

\(\vec{AP}\)=𝜆\(\vec{b}\)

Also, we can write vector AP in the following manner:

\(\vec{AP}\)=\(\vec{OP}\)\(\vec{OA}\)

𝜆\(\vec{b}\) =\(\vec{r}\)\(\vec{a}\)

\(\vec{a}\)=\(\vec{a}\)+𝜆\(\vec{b}\)

\(\vec{b}\)=𝑏1\(\hat{i}\)+𝑏2\(\hat{j}\) +𝑏3\(\hat{k}\)