NIMCET Previous Year Questions (PYQs) – Tangent And Normal

Curve: $x^2{y}^2-2x=4(1-y)$. Point: $(2,-2)$.

Differentiate implicitly:
${\displaystyle \frac{d}{dx}}(x^2{y}^2)-2={\displaystyle \frac{d}{dx}}(4(1-y))$
$2x{y}^2+2x^{2}y{y}^{\prime }-2=-4y^{\prime }$
$\Rightarrow y^{\prime }(2x^{2}y+4)=2-2x{y}^2$
$\Rightarrow y^{\prime }={\displaystyle \frac{1-x{y}^2}{x^{2}y+2}}$.

Slope at $(2,-2)$:
$x{y}^2=2\cdot 4=8$, $x^{2}y+2=4\cdot (-2)+2=-6$.
$m=y^{\prime }={\displaystyle \frac{1-8}{-6}}={\displaystyle \frac{-7}{-6}}={\displaystyle \frac{7}{6}}$.

Tangent line at $(2,-2)$:
$y+2={\displaystyle \frac{7}{6}}(x-2)\Rightarrow 6y=7x-26\Rightarrow \boxed{{\displaystyle y=\frac{7}{6}x-\frac{13}{3}}}$.

How to decide “does not pass through”: A point $(x_0,y_0)$ lies on the tangent iff $6y_0=7x_0-26$. If this fails, the tangent does not pass through that point.

Checks (examples): On the line: $(0,-\frac{13}{3})$, $(\frac{26}{7},0)$. Any point not satisfying $6y=7x-26$ is not on the tangent.