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

$\displaystyle \textbf{Question 1. }\text{Any tangent to the curve } y=3x^7+5x+3 \ \$
$\displaystyle \text{(a) is parallel to X-axis (b) is parallel to Y-axis} \ \$
$\displaystyle \text{(c) makes an acute angle with X-axis (d) makes an obtuse angle with Y-axis} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c) We have, } y=3x^{7}+5x+3$
$\displaystyle \text{On differentiating w.r.t. } x, \text{ we get}$
$\displaystyle \frac{dy}{dx}=21x^{6}+5$
$\displaystyle \text{Since } x^{6}\geq 0 \text{ for all } x,$
$\displaystyle \therefore 21x^{6}+5>0$
$\displaystyle \Rightarrow \frac{dy}{dx}>0$
$\displaystyle \text{Hence, the tangent makes an acute angle with the positive } x\text{-axis.}$
$\\$

$\displaystyle \textbf{Question 2. }\text{Slope of normal to } x^2+3y+y^2=5 \text{ at }(1,1)\ \ \$
$\displaystyle \text{(a) }-\frac{2}{5}\ (b)\ \frac{5}{2}\ (c)\ \frac{2}{5}\ (d)\ -\frac{5}{2} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) We have, } x^{2}+3y+y^{2}=5$
$\displaystyle \text{On differentiating w.r.t. } x, \text{ we get}$
$\displaystyle \frac{d}{dx}(x^{2})+\frac{d}{dx}(3y)+\frac{d}{dx}(y^{2})=\frac{d}{dx}(5)$
$\displaystyle \Rightarrow 2x+3\frac{dy}{dx}+2y\frac{dy}{dx}=0$
$\displaystyle \Rightarrow 2x+(3+2y)\frac{dy}{dx}=0$
$\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{2x}{3+2y}$
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$\displaystyle \textbf{Question 3. }\text{Find point on } y^2=4x+8 \text{ where } dx = dy$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required point be } (x,y)$
$\displaystyle \text{Given, } \frac{dy}{dt}=\frac{dx}{dt} \qquad \cdots(i)$
$\displaystyle \text{and } y^{2}=4x+8 \qquad \cdots(ii)$
$\displaystyle \text{On differentiating Eq. (ii) w.r.t. } t, \text{ we get}$
$\displaystyle 2y\frac{dy}{dt}=4\frac{dx}{dt}$
$\displaystyle \Rightarrow 2y=4 \quad [\because \frac{dy}{dt}=\frac{dx}{dt}]$
$\displaystyle \Rightarrow y=2$
$\displaystyle \text{On putting } y=2 \text{ in Eq. (ii), we get}$
$\displaystyle (2)^{2}=4x+8 \Rightarrow 4=4x+8$
$\displaystyle \Rightarrow x=-1$
$\displaystyle \therefore \text{Required point is } (-1,2)$
$\\$

$\displaystyle \textbf{Question 4. }\text{Line } y=6x-1 \text{ is tangent to } f(x) \text{ at } x=4 \ \$
$\displaystyle \text{(i) } f'(4)= ? \quad (ii)\ f(4)= ? \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } y=6x-1 \text{ is the tangent line at } x=4$
$\displaystyle \text{On differentiating, } \frac{dy}{dx}=6$
$\displaystyle \therefore f'(4)=6$
$\displaystyle \text{The tangent touches the function at } x=4$
$\displaystyle \Rightarrow f(4)=6(4)-1=23$
$\\$

$\displaystyle \textbf{Question 5. }\text{Find the equations of the normals to the curve } y=x^3+2x+6, \ \$
$\displaystyle \text{which are parallel to the line } x+14y+4=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, curve } y=x^{3}+2x+6 \qquad \cdots(i)$
$\displaystyle \text{and line } x+14y+4=0 \qquad \cdots(ii)$
$\displaystyle \text{On differentiating Eq. (i), } \frac{dy}{dx}=3x^{2}+2$
$\displaystyle \text{Slope of normal }=-\frac{1}{3x^{2}+2}$
$\displaystyle \text{Slope of line } x+14y+4=0 \text{ is } -\frac{1}{14}$
$\displaystyle \text{Since normal is parallel to line,}$
$\displaystyle -\frac{1}{3x^{2}+2}=-\frac{1}{14}$
$\displaystyle \Rightarrow 3x^{2}+2=14$
$\displaystyle \Rightarrow 3x^{2}=12 \Rightarrow x^{2}=4 \Rightarrow x=\pm 2$
$\displaystyle \text{From Eq. (i), at } x=2,\ y=(2)^{3}+2(2)+6=8+4+6=18$
$\displaystyle \text{and at } x=-2,\ y=(-2)^{3}+2(-2)+6=-8-4+6=-6$
$\displaystyle \therefore \text{Normal passes through } (2,18) \text{ and } (-2,-6)$
$\displaystyle \text{Slope of normal }=-\frac{1}{14}$
$\displaystyle \text{Equation at } (2,18):\ y-18=-\frac{1}{14}(x-2)$
$\displaystyle \Rightarrow 14y-252=-x+2 \Rightarrow x+14y=254$
$\displaystyle \text{At } (-2,-6):\ y+6=-\frac{1}{14}(x+2)$
$\displaystyle \Rightarrow 14y+84=-x-2 \Rightarrow x+14y=-86$
$\displaystyle \therefore \text{Normals are } x+14y=254 \text{ and } x+14y=-86$
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$\displaystyle \textbf{Question 6. }\text{Find point on } y=x^2+2x-1 \text{ where tangent is parallel to X-axis }$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } y=x^{2}+2x-1$
$\displaystyle \frac{dy}{dx}=2x+2=2(x+1)$
$\displaystyle \text{Let at } (x_{1},y_{1}) \text{ tangent is parallel to } x\text{-axis}$
$\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{(x_{1},y_{1})}=0$
$\displaystyle \Rightarrow 2(x_{1}+1)=0 \Rightarrow x_{1}=-1$
$\displaystyle y_{1}=(-1)^{2}+2(-1)-1=1-2-1=-2$
$\displaystyle \therefore (x_{1},y_{1})=(-1,-2)$
$\\$