Question

If the tangents are drawn at the points of intersection of the curves $3x^2+y^2-4y=0$ and $3x^2-y-2=0,$ then the slope of tangents to the first curve is/are

• A. $3$
• B. $\sqrt{3}$
• C. $-2\sqrt{3}$
• D. $-3$

Answer 1

Answer: (A), (C), (D)

$-3$

Given curves : $3x^2+y^2-4y=0$ and $3x^2-y-2=0$
For point of intersection
$3x^2+y^2-4y=3x^2-y-2$
$\Rightarrow y^2-3y+2=0$
$\Rightarrow y=1,2$
If $y=1,$ then
$3x^2+1-4=0$
$\Rightarrow 3x^2=3,x=\pm 1$
So, intersection points are $(-1,1)$ and $(1,1)$
If $y=2,$ then
$3x^2+4-8=0$
$\Rightarrow x=\pm \frac{2}{\sqrt{3}}$
So, intersection points are $(\frac{2}{\sqrt{3}},3)$ and $(-\frac{2}{\sqrt{3}},3)$

Now, first curve $3x^2+y^2-4y=0$
Differentianting both sides w.r.t. $x$
$\Rightarrow \frac{dy}{dx}=\frac{6x}{4-2y}$
Slope of tangent at $(1,1)$
$\Rightarrow \frac{dy}{dx}{|}_{(1,1)}=3$
Slope of tangent at $(-1,1)$
$\Rightarrow \frac{dy}{dx}{|}_{(-1,1)}=-3$
Slope of tangent at $(\frac{2}{\sqrt{3}},3)$
$\Rightarrow \frac{dy}{dx}{|}_{(2/\sqrt{3},1)}=2\sqrt{3}$
Slope of tangent at $(-\frac{2}{\sqrt{3}},3)$
$\Rightarrow \frac{dy}{dx}{|}_{(-2/\sqrt{3},1)}=-2\sqrt{3}$