Class 12: Tanget and Normals – Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \textbf{Question 1: }\text{Show that the tangents to the curve } y=2x^3-3 \text{ at the points where } \\ x=2 \text{ and } x=-2 \text{ are parallel.} \hspace{5.0cm} [\text{CBSE 20011}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the curve is } y=2x^3-3.$
$\displaystyle \text{Differentiating with respect to } x,\ \text{we get}$
$\displaystyle \frac{dy}{dx}=6x^2.$
$\displaystyle \text{Now, } m_1=\text{(Slope of the tangent at } x=2)=\left(\frac{dy}{dx}\right)_{x=2}=6\times (2)^2=24$
$\displaystyle \text{and, } m_2=\text{(Slope of the tangent at } x=-2)=\left(\frac{dy}{dx}\right)_{x=-2}=6\times (-2)^2=24.$
$\displaystyle \text{Clearly, } m_1=m_2.$
$\displaystyle \text{Thus, the tangents to the given curve at the points where } x=2 \text{ and } x=-2 \text{ are parallel.}$

$\displaystyle \textbf{Question 2: }\text{Prove that the tangents to the curve } y=x^2-5x+6 \text{ at the points } (2,0) \\ \text{ and } (3,0) \text{ are at right angles.} \hspace{5.0cm} [\text{CBSE 2004}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the curve is } y=x^2-5x+6.$
$\displaystyle \text{Differentiating with respect to } x,\ \text{we get}$
$\displaystyle \frac{dy}{dx}=2x-5.$
$\displaystyle \text{Now, } m_1=\text{Slope of the tangent at } (2,0)=\left(\frac{dy}{dx}\right)_{(2,0)}=2\times 2-5=-1$
$\displaystyle \text{and, } m_2=\text{Slope of the tangent at } (3,0)=\left(\frac{dy}{dx}\right)_{(3,0)}=2\times 3-5=1.$
$\displaystyle \text{Clearly, } m_1m_2=-1\times 1=-1.$
$\displaystyle \text{Thus, the tangents to the given curve at } (2,0) \text{ and } (3,0) \text{ are at right angles.}$

$\displaystyle \textbf{Question 3: }\text{Find the points on the curve } y=x^3 \text{ at which the slope of the tangent is} \\ \text{equal to the } y\text{-coordinate of the point.} \hspace{5.0cm} [\text{CBSE 2010}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required point on the curve } y=x^3 \text{ be } P(x_1,y_1).$
$\displaystyle \text{We have, } y=x^3 \Rightarrow \frac{dy}{dx}=3x^2 \Rightarrow \left(\frac{dy}{dx}\right)_{(x_1,y_1)}=3x_1^2$
$\displaystyle \text{It is given that:}$
$\displaystyle \text{Slope of the tangent at } P(x_1,y_1)=\text{Ordinate of } P(x_1,y_1)$
$\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{(x_1,y_1)}=y_1$
$\displaystyle \Rightarrow 3x_1^2=y_1$
$\displaystyle \Rightarrow 3x_1^2=x_1^3$
$\displaystyle \Rightarrow x_1^2(x_1-3)=0 \Rightarrow x_1=0,\ x_1=3$
$\displaystyle \text{Since } (x_1,y_1) \text{ lies on } y=x^3.\ \text{Therefore, } y_1=x_1^3$
$\displaystyle \therefore\ x_1=0 \Rightarrow y_1=0 \text{ and } x_1=3 \Rightarrow y_1=3^3=27$
$\displaystyle \text{Hence, required points are } (0,0) \text{ and } (3,27).$

$\displaystyle \textbf{Question 4: }\text{Find the equations of the tangent and normal to the parabola } \\ y^2=4ax \text{ at the point } (at^2,2at). \hspace{5.0cm} [\text{CBSE 2013}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the given curve is } y^2=4ax\ \ \text{...(i)}$
$\displaystyle \text{Differentiating (i) with respect to } x,\ \text{we get}$
$\displaystyle 2y\frac{dy}{dx}=4a \Rightarrow \frac{dy}{dx}=\frac{2a}{y} \Rightarrow \left(\frac{dy}{dx}\right)_{(at^2,2at)}=\frac{2a}{2at}=\frac{1}{t}$
$\displaystyle \text{So, the equation of the tangent at } (at^2,2at) \text{ is}$
$\displaystyle y-2at=\left(\frac{dy}{dx}\right)_{(at^2,2at)}(x-at^2)$
$\displaystyle \Rightarrow y-2at=\frac{1}{t}(x-at^2) \Rightarrow ty=x+at^2$
$\displaystyle \text{The equation of the normal at } (at^2,2at) \text{ is}$
$\displaystyle y-2at=-\frac{1}{\left(\frac{dy}{dx}\right)_{(at^2,2at)}}(x-at^2)$
$\displaystyle \Rightarrow y-2at=-t(x-at^2)$
$\displaystyle \Rightarrow y-2at=-tx+at^3 \Rightarrow y+tx=2at+at^3$

$\displaystyle \textbf{Question 5: }\text{Find the equation of the tangent line to the curve } x=1-\cos\theta, \\ \ y=\theta-\sin\theta \text{ at } \theta=\frac{\pi}{4}. \hspace{5.0cm} [\text{CBSE 2004}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Putting } \theta=\frac{\pi}{4} \text{ in } x=1-\cos\theta \text{ and } y=\theta-\sin\theta,\ \text{we get}$
$\displaystyle x=1-\cos\frac{\pi}{4}=1-\frac{1}{\sqrt{2}} \text{ and } y=\frac{\pi}{4}-\sin\frac{\pi}{4}=\frac{\pi}{4}-\frac{1}{\sqrt{2}}.$
$\displaystyle \text{So, coordinates of the point of contact are } \left(1-\frac{1}{\sqrt{2}},\ \frac{\pi}{4}-\frac{1}{\sqrt{2}}\right).$
$\displaystyle \text{Now, } x=1-\cos\theta \text{ and } y=\theta-\sin\theta$
$\displaystyle \Rightarrow \frac{dx}{d\theta}=\sin\theta \text{ and } \frac{dy}{d\theta}=1-\cos\theta$
$\displaystyle \therefore\ \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{1-\cos\theta}{\sin\theta}$
$\displaystyle \text{At } \theta=\frac{\pi}{4},\ \text{we get}$
$\displaystyle \frac{dy}{dx}=\frac{1-\cos\frac{\pi}{4}}{\sin\frac{\pi}{4}}=\frac{1-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=\sqrt{2}-1$
$\displaystyle \text{So, the equation of the tangent line at } \theta=\frac{\pi}{4} \text{ is}$
$\displaystyle y-\left(\frac{\pi}{4}-\frac{1}{\sqrt{2}}\right)=(\sqrt{2}-1)\left\{x-\left(1-\frac{1}{\sqrt{2}}\right)\right\}$
$\displaystyle \text{or, } (\sqrt{2}-1)x-y=2(\sqrt{2}-1)-\frac{\pi}{4}.$

$\displaystyle \textbf{Question 6: }\text{Find the equations of the tangent and the normal at the point } 't' \\ \text{ on the curve } x=a\sin^3 t,\ y=b\cos^3 t. \hspace{5.0cm} [\text{CBSE 2010, 2014}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle x=a\sin^3 t \text{ and } y=b\cos^3 t$
$\displaystyle \Rightarrow \frac{dx}{dt}=3a\sin^2 t\cos t \text{ and, } \frac{dy}{dt}=-3b\cos^2 t\sin t$
$\displaystyle \therefore\ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-3b\cos^2 t\sin t}{3a\sin^2 t\cos t}=-\frac{b}{a}\frac{\cos t}{\sin t}$
$\displaystyle \text{So, the equation of the tangent at the point } 't' \text{ is}$
$\displaystyle y-b\cos^3 t=\left(\frac{dy}{dx}\right)(x-a\sin^3 t)$
$\displaystyle \text{or, } y-b\cos^3 t=-\frac{b}{a}\frac{\cos t}{\sin t}(x-a\sin^3 t)$
$\displaystyle \text{or, } bx\cos t+ay\sin t=ab\sin t\cos t$
$\displaystyle \text{The equation of the normal at the point } 't' \text{ is}$
$\displaystyle y-b\cos^3 t=-\frac{1}{\left(\frac{dy}{dx}\right)}(x-a\sin^3 t)$
$\displaystyle \text{or, } y-b\cos^3 t=-\frac{1}{-\frac{b}{a}\frac{\cos t}{\sin t}}(x-a\sin^3 t)$
$\displaystyle \text{or, } ax\sin t-by\cos t=a^2\sin^4 t-b^2\cos^4 t$

$\displaystyle \textbf{Question 7: }\text{Show that the line } \frac{x}{a}+\frac{y}{b}=1 \text{ touches the curve } y=be^{-x/a} \text{ at the} \\ \text{point where it crosses the } y\text{-axis.} \hspace{5.0cm} [\text{CBSE 2005, 2007}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the given curve is } y=be^{-x/a}\ \ \text{...(i)}$
$\displaystyle \text{It crosses } y\text{-axis at the point, where } x=0.\ \text{Putting } x=0 \text{ in (i), we get: } y=be^0=b$
$\displaystyle \text{So, the point of contact is } (0,b).$
$\displaystyle \text{Differentiating (i) with respect to } x,\ \text{we get}$
$\displaystyle \frac{dy}{dx}=be^{-x/a}\frac{d}{dx}\left(-\frac{x}{a}\right)\Rightarrow \frac{dy}{dx}=-\frac{b}{a}e^{-x/a}\Rightarrow \left(\frac{dy}{dx}\right)_{(0,b)}=-\frac{b}{a}e^0=-\frac{b}{a}$
$\displaystyle \text{The equation of the tangent at } (0,b) \text{ is}$
$\displaystyle y-b=\left(\frac{dy}{dx}\right)_{(0,b)}(x-0)$
$\displaystyle \Rightarrow y-b=-\frac{b}{a}(x-0)\Rightarrow ay-ab=-bx\Rightarrow bx+ay=ab\Rightarrow \frac{x}{a}+\frac{y}{b}=1.$
$\displaystyle \text{Hence, } \frac{x}{a}+\frac{y}{b}=1 \text{ touches the curve } y=be^{-x/a} \text{ at the point where it crosses the axis of } y.$

$\displaystyle \textbf{Question 8: }\text{Find the equations of the tangent and the normal to the curve } \\ y=\frac{x-7}{(x-2)(x-3)} \text{ at the point, where it cuts } x\text{-axis.} \hspace{3.0cm} [\text{CBSE 2010}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the given curve is}$
$\displaystyle y(x-2)(x-3)-x+7=0\ \ \text{...(i)}$
$\displaystyle \text{This cuts the } x\text{-axis at the point, where } y=0.\ \text{Putting } y=0 \text{ in (i), we get}$
$\displaystyle -x+7=0 \Rightarrow x=7$
$\displaystyle \text{So, the point of contact is } (7,0).$
$\displaystyle \text{Differentiating (i) with respect to } x,\ \text{we get}$
$\displaystyle \frac{dy}{dx}(x-2)(x-3)+y(2x-5)-1=0\ \ \text{...(ii)}$
$\displaystyle \text{Putting } x=7 \text{ and } y=0 \text{ in (ii), we get}$
$\displaystyle \left(\frac{dy}{dx}\right)_{(7,0)}(7-2)(7-3)-1=0 \Rightarrow \left(\frac{dy}{dx}\right)_{(7,0)}=\frac{1}{20}$
$\displaystyle \text{So, the equation of the tangent at } (7,0) \text{ is}$
$\displaystyle y-0=\left(\frac{dy}{dx}\right)_{(7,0)}(x-7)\Rightarrow y-0=\frac{1}{20}(x-7)\Rightarrow x-20y-7=0$
$\displaystyle \text{The equation of the normal at } (7,0) \text{ is}$
$\displaystyle y-0=-\frac{1}{\left(\frac{dy}{dx}\right)_{(7,0)}}(x-7)\Rightarrow y-0=-20(x-7)\Rightarrow 20x+y-140=0$

$\displaystyle \textbf{Question 9: }\text{Find the equation of the normal to the curve } x^2=4y \text{ which passes} \\ \text{through the point } (1,2). \hspace{5.0cm} [\text{CBSE 2013}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Suppose the normal at } P(x_1,y_1) \text{ on the parabola } x^2=4y \text{ passes }\text{through the point } \\ (1,2).\ \text{Since } P(x_1,y_1) \text{ lies on } x^2=4y.$
$\displaystyle \therefore\ x_1^2=4y_1\ \ \text{...(i)}$
$\displaystyle \text{The equation of the curve is } x^2=4y.$
$\displaystyle \text{Differentiating with respect to } x,\ \text{we get}$
$\displaystyle 2x=4\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{x}{2}\Rightarrow \left(\frac{dy}{dx}\right)_P=\frac{x_1}{2}$
$\displaystyle \text{The equation of the normal at } P(x_1,y_1) \text{ is}$
$\displaystyle y-y_1=-\frac{1}{\left(\frac{dy}{dx}\right)_P}(x-x_1)\Rightarrow y-y_1=-\frac{2}{x_1}(x-x_1)\ \ \text{...(ii)}$
$\displaystyle \text{It passes through } (1,2).$
$\displaystyle \therefore\ 2-y_1=-\frac{2}{x_1}(1-x_1)\Rightarrow 2-y_1=-\frac{2}{x_1}+2\Rightarrow x_1y_1=2\ \ \text{...(iii)}$
$\displaystyle \text{Eliminating } y_1 \text{ between (i) and (iii), we obtain}$
$\displaystyle \frac{x_1^2}{4}\cdot x_1=2\Rightarrow \frac{x_1^3}{4}=2\Rightarrow x_1^3=8\Rightarrow x_1=2$
$\displaystyle \text{Putting } x_1=2 \text{ in (ii), we get } y_1=1.$
$\displaystyle \text{Putting the values of } x_1 \text{ and } y_1 \text{ in (ii), we get}$
$\displaystyle y-1=-1(x-2)\Rightarrow x+y-3=0,\ \text{which is the required equation of the normal.}$

$\displaystyle \textbf{Question 10: }\text{For the curve } y=4x^3-2x^5 \text{ find all points at which the tangent} \\ \text{passes through the origin.} \hspace{5.0cm} [\text{CBSE 2013}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } (x_1,y_1) \text{ be the required point on } y=4x^3-2x^5.\ \text{Then,}$
$\displaystyle y_1=4x_1^3-2x_1^5\ \ \text{...(i)}$
$\displaystyle \text{The equation of the given curve is } y=4x^3-2x^5.$
$\displaystyle \text{Differentiating with respect to } x,\ \text{we get}$
$\displaystyle \frac{dy}{dx}=12x^2-10x^4\Rightarrow \left(\frac{dy}{dx}\right)_{(x_1,y_1)}=12x_1^2-10x_1^4$
$\displaystyle \text{So, the equation of the tangent at } (x_1,y_1) \text{ is}$
$\displaystyle y-y_1=\left(\frac{dy}{dx}\right)_{(x_1,y_1)}(x-x_1)\Rightarrow y-y_1=(12x_1^2-10x_1^4)(x-x_1)$
$\displaystyle \text{This passes through the origin. Therefore,}$
$\displaystyle 0-y_1=(12x_1^2-10x_1^4)(0-x_1)\Rightarrow y_1=12x_1^3-10x_1^5\ \ \text{...(ii)}$
$\displaystyle \text{Subtracting (ii) from (i), we get}$
$\displaystyle 0=-8x_1^3+8x_1^5\Rightarrow 8x_1^3(x_1^2-1)=0\Rightarrow x_1=0 \text{ or, } x_1=\pm 1$
$\displaystyle \text{When } x_1=0 \Rightarrow y_1=0\ \ [\text{Using (ii)}]$
$\displaystyle \text{When } x_1=1 \Rightarrow y_1=12-10=2\ \ [\text{Using (ii)}]$
$\displaystyle \text{When } x_1=-1 \Rightarrow y_1=-12+10=-2\ \ [\text{Using (ii)}]$
$\displaystyle \text{Hence, the required points are } (0,0),\ (1,2) \text{ and } (-1,-2).$

$\displaystyle \textbf{Question 11: }\text{Prove that all normals to the curve } x=a\cos t+at\sin t, \\ y=a\sin t-at\cos t \text{ are at a distance } a \text{ from the origin.} \hspace{3.0cm} [\text{CBSE 2013}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equations of the curve are:}$
$\displaystyle x=a\cos t+at\sin t \text{ and } y=a\sin t-at\cos t$
$\displaystyle \Rightarrow \frac{dx}{dt}=at\cos t \text{ and } \frac{dy}{dt}=at\sin t$
$\displaystyle \therefore\ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{at\sin t}{at\cos t}=\tan t$
$\displaystyle \text{The equation of the normal at any point } t \text{ is given by}$
$\displaystyle y-(a\sin t-at\cos t)=-\frac{1}{\frac{dy}{dx}}\{x-(a\cos t+at\sin t)\}$
$\displaystyle \Rightarrow y-(a\sin t-at\cos t)=-\frac{1}{\tan t}\{x-(a\cos t+at\sin t)\}$
$\displaystyle \Rightarrow y-(a\sin t-at\cos t)=-\frac{\cos t}{\sin t}\{x-(a\cos t+at\sin t)\}$
$\displaystyle \Rightarrow y\sin t-(a\sin^2 t-at\sin t\cos t)=-x\cos t+a\cos^2 t+at\sin t\cos t$
$\displaystyle \Rightarrow x\cos t+y\sin t=a\ \ \text{...(i)}$
$\displaystyle \therefore\ \text{Length of the perpendicular from the origin to (i) } \\ =\frac{|0\cos t+0\sin t-a|}{\sqrt{\cos^2 t+\sin^2 t}}=a$
$\displaystyle \text{Hence, all normals to the given curve are at a distance } 'a' \text{ from the origin.}$

$\displaystyle \textbf{Question 12: }\text{Find all the tangents to the curve } y=\cos(x+y), \\ -2\pi\le x\le 2\pi \text{ that are parallel to the line } x+2y=0. \hspace{3.0cm} [\text{CBSE 2016, 2017}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the point of contact of one of the tangents be } (x_1,y_1).\ \text{Then, } (x_1,y_1) \text{ lies on } y=\cos(x+y)$
$\displaystyle \therefore\ y_1=\cos(x_1+y_1)\ \ \text{...(i)}$
$\displaystyle \text{Since the tangents are parallel to the line } x+2y=0.\ \text{Therefore,}$
$\displaystyle \text{Slope of the tangent at } (x_1,y_1)=\text{(Slope of the line } x+2y=0)$
$\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{(x_1,y_1)}=-\frac{1}{2}$
$\displaystyle \text{The equation of the curve is } y=\cos(x+y).$
$\displaystyle \text{Differentiating with respect to } x,\ \text{we get}$
$\displaystyle \frac{dy}{dx}=-\sin(x+y)\left(1+\frac{dy}{dx}\right)$
$\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{(x_1,y_1)}=-\sin(x_1+y_1)\left\{1+\left(\frac{dy}{dx}\right)_{(x_1,y_1)}\right\}$
$\displaystyle \Rightarrow -\frac{1}{2}=-\sin(x_1+y_1)\left(1-\frac{1}{2}\right)$
$\displaystyle \Rightarrow \sin(x_1+y_1)=1\ \ \text{...(ii)}$
$\displaystyle \text{Squaring (i) and (ii) and then adding, we get}$
$\displaystyle \cos^2(x_1+y_1)+\sin^2(x_1+y_1)=y_1^2+1 \Rightarrow 1=y_1^2+1 \Rightarrow y_1=0$
$\displaystyle \text{Putting } y_1=0 \text{ in (i) and (ii), we get}$
$\displaystyle \cos x_1=0 \text{ and } \sin x_1=1 \Rightarrow x_1=\frac{\pi}{2},-\frac{3\pi}{2}\ \ \ [\therefore\ -2\pi\le x_1\le 2\pi]$
$\displaystyle \text{Hence, the points of contact are } \left(\frac{\pi}{2},0\right) \text{ and } \left(-\frac{3\pi}{2},0\right).$
$\displaystyle \text{The slope of the tangent is } \left(-\frac{1}{2}\right).\ \text{Therefore, equations of tangents at } \left(\frac{\pi}{2},0\right) \text{ and } \left(-\frac{3\pi}{2},0\right) \text{ are}$
$\displaystyle y-0=-\frac{1}{2}\left(x-\frac{\pi}{2}\right) \text{ and } y-0=-\frac{1}{2}\left(x+\frac{3\pi}{2}\right) \text{ respectively}$
$\displaystyle \text{or, } 2x+4y-\pi=0 \text{ and } 2x+4y+3\pi=0 \text{ respectively.}$

$\displaystyle \textbf{Question 13: }\text{Find the angle between the parabolas } y^2=4ax \text{ and } x^2=4by \text{ at their} \\ \text{point of intersection other than the origin.} \hspace{5.0cm} [\text{CBSE 2016}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equations of two parabolas are } y^2=4ax \text{ and } x^2=4by.$
$\displaystyle \text{Now, } x^2=4by \Rightarrow y=\frac{x^2}{4b}.$
$\displaystyle \text{Substituting this value of } y \text{ in } y^2=4ax,\ \text{we get}$
$\displaystyle \left(\frac{x^2}{4b}\right)^2=4ax$
$\displaystyle \Rightarrow x^4-64ab^2x=0 \Rightarrow x(x^3-64ab^2)=0 \Rightarrow x=0,\ x^3=64ab^2$
$\displaystyle \Rightarrow x=0,\ x=4a^{1/3}b^{2/3}$
$\displaystyle \text{Putting } x=0 \text{ and } x=4a^{1/3}b^{2/3} \text{ successively in } y=\frac{x^2}{4b},\ \text{we get } y=0 \text{ and } y=4a^{2/3}b^{1/3} \text{ respectively.}$
$\displaystyle \text{Thus, the two curves intersect at } P\left(4a^{1/3}b^{2/3},4a^{2/3}b^{1/3}\right) \text{ other than the origin } O(0,0).$
$\displaystyle \text{Now, } y^2=4ax \text{ and } x^2=4by$
$\displaystyle \Rightarrow 2y\frac{dy}{dx}=4a \text{ and } 2x=4b\frac{dy}{dx}\ \ \text{[Differentiating both with respect to } x]$
$\displaystyle \Rightarrow \frac{dy}{dx}=\frac{2a}{y} \text{ and } \frac{dy}{dx}=\frac{x}{2b}$
$\displaystyle \Rightarrow m_1=\left(\frac{dy}{dx}\right)_P=\frac{2a}{4a^{2/3}b^{1/3}}=\frac{1}{2}\left(\frac{a}{b}\right)^{1/3}$
$\displaystyle \text{and } m_2=\left(\frac{dy}{dx}\right)_P=\frac{4a^{1/3}b^{2/3}}{2b}=2\left(\frac{a}{b}\right)^{1/3}$
$\displaystyle \text{Let } \theta \text{ be the angle between the tangents to the parabolas } y^2=4ax \text{ and } x^2=4by \text{ at } P.\ \text{Then,}$
$\displaystyle \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$
$\displaystyle \Rightarrow \tan\theta=\left|\frac{\frac{1}{2}\left(\frac{a}{b}\right)^{1/3}-2\left(\frac{a}{b}\right)^{1/3}}{1+\frac{1}{2}\left(\frac{a}{b}\right)^{1/3}\cdot 2\left(\frac{a}{b}\right)^{1/3}}\right|$
$\displaystyle \Rightarrow \tan\theta=\left|\frac{-\frac{3}{2}\left(\frac{a}{b}\right)^{1/3}}{1+\left(\frac{a}{b}\right)^{2/3}}\right|=\frac{3a^{1/3}b^{1/3}}{2(a^{2/3}+b^{2/3})}$
$\displaystyle \therefore\ \theta=\tan^{-1}\left\{\frac{3(ab)^{1/3}}{2(a^{2/3}+b^{2/3})}\right\}.$

$\displaystyle \textbf{Question 14: }\text{Show that the curves } x=y^2 \text{ and } xy=k \text{ cut at right angles,} \\ \text{if } 8k^2=1. \hspace{5.0cm} [\text{CBSE 2004, 2005, 2013}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given curves are}$
$\displaystyle x=y^2\ \ \text{...(i)}$
$\displaystyle \text{and,}\ \ xy=k\ \ \text{...(ii)}$
$\displaystyle \text{From (i), we obtain } x=y^2.\ \text{Putting this value of } x \text{ in (ii), we obtain}$
$\displaystyle y^3=k \Rightarrow y=k^{1/3}$
$\displaystyle \text{Putting } y=k^{1/3} \text{ in (i), we get } x=k^{2/3}$
$\displaystyle \text{So, the two curves intersect at the point } P(k^{2/3},k^{1/3}).$
$\displaystyle \text{Differentiating (i) with respect to } x,\ \text{we get}$
$\displaystyle 1=2y\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{1}{2y}\Rightarrow m_1=\left(\frac{dy}{dx}\right)_P=\frac{1}{2k^{1/3}}$
$\displaystyle \text{Differentiating (ii) with respect to } x,\ \text{we get}$
$\displaystyle y+x\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{y}{x}\Rightarrow m_2=\left(\frac{dy}{dx}\right)_P=-\frac{k^{1/3}}{k^{2/3}}=-\frac{1}{k^{1/3}}$
$\displaystyle \text{For the curves (i) and (ii) to cut at right angles at } P,\ \text{we must have}$
$\displaystyle m_1m_2=-1\Rightarrow \frac{1}{2k^{1/3}}\cdot\left(-\frac{1}{k^{1/3}}\right)=-1\Rightarrow 2k^{2/3}=1$
$\displaystyle \Rightarrow (2k^{2/3})^3=1\Rightarrow 8k^2=1.$

$\displaystyle \textbf{Question 15: }\text{Find the values of } p \text{ for which the curves } x^2=9p(9-y) \text{ and } \\ x^2=p(y+1) \text{ cut each other at right angles.} \hspace{5.0cm} [\text{CBSE 2015}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equations of the given curves are}$
$\displaystyle x^2=9p(9-y)\ \ \text{...(i)}$
$\displaystyle \text{and,}\ \ x^2=p(y+1)\ \ \text{...(ii)}$
$\displaystyle \text{To find the coordinates of the point(s) of intersection of (i) and (ii), we solve the two} \\ \text{equations simultaneously. On eliminating } x^2,\ \text{we obtain}$
$\displaystyle 9p(9-y)=p(y+1)\Rightarrow 81-9y=y+1\Rightarrow 10y=80\Rightarrow y=8$
$\displaystyle \text{Putting } y=8 \text{ in (i) or (ii), we obtain}$
$\displaystyle x^2=9p\Rightarrow x=\pm 3\sqrt{p}$
$\displaystyle \text{Thus, curves (i) and (ii) intersect at } P(3\sqrt{p},8) \text{ and } Q(-3\sqrt{p},8).$
$\displaystyle \text{Differentiating (i) and (ii) with respect to } x,\ \text{we obtain}$
$\displaystyle 2x=-9p\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=-\frac{2x}{9p}\Rightarrow m_1=\left(\frac{dy}{dx}\right)_{C_1}=-\frac{2\times 3\sqrt{p}}{9p}=-\frac{2}{3\sqrt{p}}$
$\displaystyle \text{Differentiating (ii) with respect to } x,\ \text{we obtain}$
$\displaystyle 2x=p\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{2x}{p}\Rightarrow m_2=\left(\frac{dy}{dx}\right)_{C_2}=\frac{2(3\sqrt{p})}{p}=\frac{6}{\sqrt{p}}$
$\displaystyle \text{If curves (i) and (ii) cut each other at } P \text{ at right angles, then}$
$\displaystyle m_1m_2=-1\Rightarrow -\frac{2}{3\sqrt{p}}\cdot\frac{6}{\sqrt{p}}=-1\Rightarrow p=4$
$\displaystyle \text{Similarly, by using the condition of orthogonality of the curves at } Q,\ \text{we obtain } \\ p=4.$
$\displaystyle \text{Hence, the two curves cut each other at right angles, if } p=4.$

$\displaystyle \textbf{Question 16: }\text{Show that the curves } xy=a^2 \text{ and } x^2+y^2=2a^2 \text{ touch} \\ \text{each other.} \hspace{5.0cm} [\text{CBSE 2002}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given curves are}$
$\displaystyle xy=a^2\ \ \text{...(i)}$
$\displaystyle x^2+y^2=2a^2\ \ \text{...(ii)}$
$\displaystyle \text{From (i), we get } y=\frac{a^2}{x}.\ \text{Substituting this value of } y \text{ in equation (ii), we get}$
$\displaystyle x^2+\frac{a^4}{x^2}=2a^2\Rightarrow x^4-2a^2x^2+a^4=0\Rightarrow (x^2-a^2)^2=0\Rightarrow x=\pm a$
$\displaystyle \text{From (i), we get}$
$\displaystyle y=a \text{ for } x=a \text{ and, } y=-a \text{ for } x=-a.$
$\displaystyle \text{Thus, the two curves intersect at } P(a,a) \text{ and } Q(-a,-a).$
$\displaystyle \text{Differentiating both sides of curve (i) with respect to } x,\ \text{we get}$
$\displaystyle x\frac{dy}{dx}+y=0\Rightarrow \frac{dy}{dx}=-\frac{y}{x}\ \ \text{...(iii)}$
$\displaystyle \text{Differentiating both sides of curve (ii) with respect to } x,\ \text{we get}$
$\displaystyle 2x+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{x}{y}\ \ \text{...(iv)}$
$\displaystyle \text{Angle of intersection at } P(a,a):\ \text{Substituting } x=a,\ y=a \text{ in (iii) and (iv), we get}$
$\displaystyle \left(\frac{dy}{dx}\right)_{C_1}=-\frac{a}{a}=-1 \text{ and, } \left(\frac{dy}{dx}\right)_{C_2}=-\frac{a}{a}=-1$
$\displaystyle \text{Clearly, } \left(\frac{dy}{dx}\right)_{C_1}=\left(\frac{dy}{dx}\right)_{C_2}\ \text{at } P.\ \text{So, the two curves touch each other at } P.$
$\displaystyle \text{Similarly, it can be seen that the two curve touch each other at } Q.$