Question

The curve $x^2 - xy + y^2 =27$ has tangents parallel to $x-axis$ at

• A. $(-3, - 6) & (3, - 6)$
• B. $(3, 6),& (-3, - 6)$
• C. $(-3,6) & (-3, -6)$
• D. $(3, -6) & (-3, 6)$

Answer 1

Answer: (B)

Given equation of curve is
$x^2 - xy + y^2 = 27 $ ... (i)
Taking derivative w.r.t. $'x'$ ori both sides
$\Rightarrow 2x -x \frac{dy}{dx} - y + 2y \frac{dy}{dx} = 0 $
$\Rightarrow \:\:\: \frac{dy}{dx} \left(2y -x\right)=y -2x $
$\Rightarrow \frac{dy}{dx} = \frac{y-2x}{2y-x} $
Since, curve has tangent parallel to x-axis
$\therefore $ slope of tangent= 0
$\Rightarrow \frac{dy}{dx} = 0 \Rightarrow \:\:\:\: \frac{y-2x}{2y-x} = 0 $
$\Rightarrow y -2x$ .....(ii)
\ Now solving (i) and (ii) we get,
$ x^{2} -2x^{2}+4x^{2} =27 $
$\Rightarrow 3x^{2}=27 \Rightarrow x = \pm3$
For $x = 3, y = 6$ and $x = -3, y = -6$
$\therefore $ Points are (3, 6) and (-3, -6)