Question

The point(s) on the curve where tangents to the curve $y^2-2x^3-4y+8=0$ passes through $(1,2)$ is

• A. $(2,3\pm 2\sqrt{3})$
• B. $(2,2\pm 2\sqrt{3})$
• C. $(3,\pm 2\sqrt{3})$
• D. $(3,\pm 3\sqrt{3})$

Answer 1

The correct option is B $(2,2\pm 2\sqrt{3})$

$y^2-2x^3-4y+8=0$
Let a tangent is drawn to the curve at point $Q(\alpha ,\beta )$ on the curve which passes through $P(1,2)$.
Differentiating w.r.t. $x$, we get
$2y\frac{dy}{dx}-6x^2-4\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{3x^2}{y-2}$
Now, slope of line $PQ={\left(\frac{dy}{dx}\right)}_{(\alpha ,\beta )}$
$\Rightarrow \frac{\beta -2}{\alpha -1}=\frac{3{\alpha }^2}{\beta -2}$
$\Rightarrow (\beta -2{)}^2=3{\alpha }^2(\alpha -1)\cdots (i)$
Also $(\alpha ,\beta )$ satisfies the equation of the curve.
$\Rightarrow {\beta }^2-2{\alpha }^3-4\beta +8=0$ or $(\beta -2{)}^2=2{\alpha }^3-4\cdots (ii)$
From equations $\left(i\right)$ and $\left(ii\right)$, $3{\alpha }^2(\alpha -1)=2{\alpha }^3-4$
$\Rightarrow {\alpha }^3-3{\alpha }^2+4=0$
$\Rightarrow (\alpha -2)({\alpha }^2-\alpha -2)=0$
$\Rightarrow (\alpha -2{)}^2(\alpha +1)=0$
When $\alpha =2,(\beta -2{)}^2=12\Rightarrow \beta =2\pm 2\sqrt{3}$
When $\alpha =-1,(\beta -2{)}^2=-6$ (not possible)
$\Rightarrow (\alpha ,\beta )\equiv (2,2\pm 2\sqrt{3})$