NIMCET Tangent And Normal Previous Year Questions (PYQs)

Curve: \(x^2y^2-2x=4(1-y)\). Point: \((2,-2)\).

Differentiate implicitly:
\(\dfrac{d}{dx}(x^2y^2)-2=\dfrac{d}{dx}\big(4(1-y)\big)\)
\(2xy^2+2x^2y\,y'-2=-4y'\)
\(\Rightarrow y'\,(2x^2y+4)=2-2xy^2\)
\(\Rightarrow y'=\dfrac{1-xy^2}{x^2y+2}\).

Slope at \((2,-2)\):
\(xy^2=2\cdot4=8\), \(x^2y+2=4\cdot(-2)+2=-6\).
\(m=y'=\dfrac{1-8}{-6}=\dfrac{-7}{-6}=\dfrac{7}{6}\).

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

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: \((0,-\tfrac{13}{3})\), \((\tfrac{26}{7},0)\). Any point not satisfying \(6y=7x-26\) is not on the tangent.

Given curves are:

$y=(x-2)^2$

and

$y=-4+6x-x^2$

At the point of intersection,

$(x-2)^2=-4+6x-x^2$

$x^2-4x+4=-4+6x-x^2$

$2x^2-10x+8=0$

$x^2-5x+4=0$

$(x-1)(x-4)=0$

So,

$x=1$ or $x=4$

Now, slopes of the curves are:

For $y=(x-2)^2$,

$m_1=\frac{dy}{dx}=2(x-2)$

For $y=-4+6x-x^2$,

$m_2=\frac{dy}{dx}=6-2x$

At $x=1$,

$m_1=2(1-2)=-2$

$m_2=6-2(1)=4$

Angle between two curves is given by:

$\tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|$

So,

$\tan\theta=\left|\frac{4-(-2)}{1+(-2)(4)}\right|$

$=\left|\frac{6}{1-8}\right|$

$=\left|\frac{6}{-7}\right|$

$=\frac{6}{7}$

Therefore,

$\theta=\tan^{-1}\left(\frac{6}{7}\right)$